Free Edge Lengths in Plane Graphs

Transcription

1 Free Edge Lengths in Plane Graphs Zachary Ael Roert Connelly Sarah Eisenstat Radoslav Fulek Filip Morić Yoshio Okamoto Tior Szaó Csaa D. Tóth Astract We study the impact of metric constraints on the realizaility of planar graphs. Let G e a sugraph of a planar graph H (where H is the host of G). The graph G is free in H if for every choice of positive lengths for the edges of G, the host H has a planar straight-line emedding that realizes these lengths; and G is extrinsically free in H if all constraints on the edge lengths of G depend on G only, irrespective of additional edges of the host H. We characterize the planar graphs G that are free in every host H, G H, and all the planar graphs G that are extrinsically free in every host H, G H. The case of cycles G = C k provides a new version of the celerated carpenter s rule prolem. Even though cycles C k, k 4, are not extrinsically free in all triangulations, it turns out that nondegenerate edge lengths are always realizale, where the edge lengths are considered degenerate if the cycle can e flattened (into a line) in two different ways. Separating triangles, and separating cycles in general, play an important role in our arguments. We show that every star is free in a 4-connected triangulation (which has no separating triangle). 1 Introduction Representing graphs in Euclidean space such that some or all of the edges have given lengths has a rich history. For example, the rigidity theory of ar-and-joint frameworks, motivated y applications in mechanics, studies edge lengths that guarantee a unique (or locally unique) representation of a graph. Our primary interest lies in simple cominatorial conditions that guarantee realizations for all possile edge lengths. We highlight two well-known results similar to ours. (1) Jackson and Jordán [4, 1] gave a cominatorial characterization of graphs that are generically gloally rigid (i.e., admit unique realizations for aritrary generic edge lengths). () Connelly et al. [5] showed that a cycle C k, k 3, emedded in the plane can e continuously unfolded into a convex polygon (i.e., the configuration space of the planar emeddings of C k is connected), solving the so-called carpenter s rule prolem. We consider straight-line emeddings of planar graphs where some of the edges can have aritrary lengths. A straight-line emedding (for short, emedding) of a planar graph is a realization in the plane where the vertices are mapped to distinct points, and the edges are mapped to line segments etween the corresponding vertices such that any two edges can intersect only at a common endpoint. By Fáry s theorem [9], every planar graph admits a straight-line emedding with some edge lengths. However, it is NP-hard to decide whether a planar graph can e emedded with prescried edge lengths [7], even for planar 3-connected graphs with unit edge lengths [3], ut it is decidale in linear time for triangulations [6] and A preliminary version of this paper appeared in the Proceedings of the 13th Annual Symposium on Computational Geometry (SoCG), 014, ACM Press, pp Massachusetts Institute of Technology, Camridge, MA, USA. zael@math.mit.edu Cornell University, Ithaca, NY, USA. connelly@math.cornell.edu Massachusetts Institute of Technology, Camridge, MA, USA. seisenst@mit.edu Columia University, New York, NY, USA. radoslav.fulek@gmail.com École Polytechnique Fédérale de Lausanne, Lausanne, Switzerland. filip.moric@epfl.ch University of Electro-Communications, Tokyo, Japan. okamotoy@uec.ac.jp Freie Universität Berlin, Berlin, Germany. szao@math.fu-erlin.de California State University Northridge, Los Angeles, CA, USA. cdtoth@acm.org 1

2 1 Introduction near-triangulations [3]. Finding a straight-line emedding of a graph with prescried edge lengths involves a fine interplay etween topological, metric, cominatorial, and algeraic constraints. Determining the impact of each of these constraints is a challenging task. In this paper, we characterize the sugraphs for which the metric constraints on the straight-line emedding remain independent from any topological, cominatorial, and algeraic constraints. Such sugraphs admit aritrary positive edge lengths in an appropriate emedding of the host graph. This motivates the following definition. Definition 1 Let G = (V, E) e a sugraph of a planar graph H (the host of G). We say that G is free in H when, for every length assignment l: E R +, there is a straight-line emedding of H in which every edge e E has length l(e); G is extrinsically free in H when, for every length assignment l: E R +, if G has a straight-line emedding with edge lengths l(e), e E, then H also has a straight-line emedding in which every edge e E has length l(e). Intuitively, if G is free in H, then there is no restriction on the edge lengths of G; and if G is extrinsically free in H, then all constraints on the edge lengths depend on G alone, rather than the edges in H \G. Clearly, if G is free in H, then it is also extrinsically free in H. However, an extrinsically free sugraph G of H need not e free in H. For example, K 3 is not free in any host since the edge lengths have to satisfy the triangle inequality, ut it is an extrinsically free sugraph in K 4. It is easily verified that every sugraph with exactly two edges is free in every host (every pair of lengths can e attained y a suitale affine transformation); ut a triangle K 3 is not free in any host (due to the triangle inequality). Results. We characterize the graphs G that are free as a sugraph of every host H, G H. Theorem 1 A planar graph G = (V, E) is free in every planar host H, G H, if and only if G consists of isolated vertices and a matching, or a forest with at most 3 edges, or the disjoint union of two paths, each with edges. Separating 3- and 4-cycles in triangulations play an important role in our argument. A star is a graph G = (V, E), where V = {v, u 1,..., u k } and E = {vu 1,..., vu k }. We present the following result for stars in 4-connected triangulations. Theorem Every star is free in a 4-connected triangulation. If a graph G is free in H, then it is extrinsically free, as well. We completely classify graphs G that are extrinsically free in every host H. Theorem 3 Let G = (V, E) e a planar graph. Then G is extrinsically free in every host H, G H, if and only if G consists of isolated vertices and a forest as listed in Theorem 1 (a matching, a forest with at most 3 edges, the disjoint union of two paths, each with edges), or a triangulation, or a triangle and one additional edge (either disjoint from or incident to the triangle).

3 Free Edge Lengths in Plane Graphs 3 When G = C k is a cycle with prescried edge lengths, the realizaility of a host H, C k H, leads to a variant of the celerated carpenter s rule prolem. Even though cycles on four or more vertices are not extrinsically free, all nonrealizale length assignments are degenerate in the sense that the cycle C k, k 4, decomposes into four paths of lengths (a,, a, ) for some a, R +. Intuitively, a length assignment on a cycle C k is degenerate if C k has two noncongruent emeddings in the line (that is, in 1-dimensions) with prescried edge lengths. We show that every host H, C k H, is realizale with prescried edge lengths on C k, that is, H admits a straight-line emedding in which every edge of C k has its prescried length, if the length assignment of C k is nondegenerate. Theorem 4 Let H e a planar graph that contains a cycle C = (V, E). Let l : E R + e a length assignment such that C has a straight-line emedding with edge lengths l(e), e E. If l is nondegenerate, then H admits a straight-line emedding in which every edge e E has length l(e). Organization. Our negative results (i.e., a planar graph G is not always free) are confirmed y finding specific hosts H, G H, and length assignments that cannot e realized (Section ). We give a constructive proof that every matching is free in all planar graphs (Section 3). In fact, we prove a slightly stronger statement: the edge lengths of a matching G can e chosen aritrarily in every plane graph H with a fixed cominatorial emedding (that is, the edge lengths and the outer face can e chosen aritrarily). The key tools are edge contractions and vertex splits, reminiscent of the technique of Fáry [9]. Separating triangles pose technical difficulties: we should realize the host H even if one edge of a separating triangle has to e very short, and an edge in its interior has to e very long. Similar prolems occur when two opposite sides of a separating 4-cycle are short. We use grid emeddings and affine transformations to construct emeddings recursively for all separating 3- and 4-cycles (Section 3.). All other sugraphs listed in Theorem 1 have at most 4 edges. We show directly that they are free in every planar host (Section 4). We extend our methods to stars in 4-connected triangulations (Section 5) and extrinsically free graphs (Section 6). In Section 7 we show that for cycles with prescried edge lengths any host H is realizale if the length assignment is nondegenerate. We conclude with open prolems in Section 8. Related Prolems. As noted aove, the emeddaility prolem for planar graphs with given edge lengths is NP-hard [3, 7], ut efficiently decidale for near-triangulations [3, 6]. Patrignani [16] also showed that it is NP-hard to decide whether a straight-line emedding of a sugraph G (i.e., a partial emedding) can e extended to an emedding of a host H, G H. For curvilinear emeddings, this prolem is known as planarity testing for partially emedded graphs (PEP), which is decidale in polynomial time []. Recently, Jelínek et al. [13] gave a cominatorial characterization for PEP via a list of foridden sustructures. Sauer [17, 18] considers similar prolems in the context of structural Ramsey theory of metric emeddings: For an edge laeled graph G and a set R R + that contains the laels, he derived conditions that ensure the existence of a metric space M on V (G) that realizes the edge laels as distances etween the endpoints. Definitions. A triangulation is an edge-maximal planar graph with n 3 vertices and 3n 6 edges. Every triangulation has well-defined faces where all faces are triangles, since every triangulation is a 3-connected polyhedral graph for n 4. A near-triangulation is a 3-connected planar graph in which all faces are triangles with at most one exception (which is typically the outer face). A 3-cycle t in a near-triangulation T is called a separating triangle if the vertices of t form a 3-cut in T. A triangulation T has no separating triangles if and only if T is 4-connected. Tools from Graph Drawing. To show that a graph G = (V, E) is free in every host H, G H, we design algorithms that, for every length assignment l: E R +, construct a desired emedding of H. Our algorithms rely on several classic uilding locks developed in the graph drawing community. By Tutte s arycenter emedding method [0], every 3-connected planar graph admits a straight-line emedding in which the outer face is mapped to an aritrarily prescried convex polygon with the right

4 4 Sugraphs with Constrained Edge Lengths numer of vertices. Hong and Nagamochi [11] extended this result, and proved that every 3-connected planar graph admits a straight-line emedding in which the outer face is mapped to an aritrarily prescried star-shaped polygon with the right numer of vertices. A grid-emedding of a planar graph is an emedding in which the vertices are mapped to points in some small h w section of the integer lattice Z. For an n-vertex planar graph, the dimensions of the ounding ox are h, w O(n) [8, 19], which is the est possile [10]. The angular resolution of a straight-line emedding of a graph is the minimum angle sutended y any two adjacent edges. It is easy to see that the angular resolution of a grid emedding, where h, w O(n), is Ω(n ). By modifying an incremental algorithm y de Fraysseix et al. [8], Kurowski [14] constructed grid emeddings of n-vertex planar graphs on a 3n 3 n section of the integer lattice with angular resolution at least 3 Ω(1/n). Kurowski s algorithm 5n emeds a n-vertex triangulation T with a given face (a,, c) such that a = (0, 0), = (3n, 0) and c = ( 3n/, 3n/ ). It has the following additional property used in our argument. When vertex c is deleted from the triangulation T, we are left with a -connected graph with an outer face (a = u 1, u,..., u k = ). In Kurowski s emedding, as well as in [8], the path (a = u 1, u,..., u k = ) is x-monotone and the slope of every edge in this path is in the range ( 1, 1). Sugraphs with Constrained Edge Lengths It is clear that a triangle is not free, since the edge lengths have to satisfy the triangle inequality in every emedding (they cannot e prescried aritrarily). This simple oservation extends to all cycles. Oservation 1 No cycle is free in any planar graph. Proof. Let C e a cycle with k 3 edges in a planar graph H. If the first k 1 edges of C have unit length, then the length of the k-th edge is less than k 1 y repeated applications of the triangle inequality. Oservation Let T e a triangulation with a separating triangle ac that separates edges e 1 and e. Then the sugraph G with edge set E = {a, c, e 1, e } is not free in T. (See Fig. 1.) Proof. Since ac separates e 1 and e, in every emedding of T, one of e 1 and e lies in the interior of ac. If a and c have unit length, then all edges of ac are shorter than in every emedding (y the triangle inequality), and hence the length of e 1 or e has to e less than. Based on Oservations 1 and, we can show that most planar graphs G are not free in some appropriate triangulations T, G T. Theorem 5 Let G = (V, E) e a forest with at least 4 edges, at least two of which are adjacent, such that G is not the disjoint union of two paths P. Then there is a triangulation T that contains G as a sugraph and G is not free in T. Proof. We shall augment G to a triangulation T such that Oservation is applicale. Specifically, we find four edges, a, c, e 1, e E, such that either e 1 and e are in distinct connected components of G or the (unique) path from e 1 to e passes through a vertex in {a,, c}. If we find four such edges, then G can e triangulated such that ac is a triangle (y adding edge ac), and it separates edges e 1 and e. See Fig. 1 for examples. We distinguish several cases ased on the maximum degree (G) of G. Case 1: (G) 4. Let e a vertex of degree at least 4 in G, with incident edges a, c, e 1 and e. Then e 1 and e are in the same component of G, and the unique path etween them contains. Case : (G) = 3. Let e a vertex of degree 3, and let e 1 e an edge not incident to. If e 1 and are in the same connected component of G, then let a e the first edge of the (unique) path from to e 1 ; otherwise

5 Free Edge Lengths in Plane Graphs 5 c c c a e 1 a e 1 e a e 1 e e Figure 1: Triangulations containing a old sugraph G with edges a, c, e 1 and e. In every emedding, one of e 1 and e lies in the interior of triangle ac, and so min{l(e 1 ), l(e )} < l(a) + l(c). Left: G has four edges, two of which are adjacent. Middle: G is a star. Right: G is a path. let a e an aritrary edge incident to. Denote the other two edges incident to y c and e. This ensures that if e 1 and e are in the same component of G, the unique path etween them contains. Case 3: (G) =. If G contains a path with 4 edges, then let the edges of the path e (e 1, a, c, e ). Now the (unique) path etween e 1 and e clearly contains a,, and c, so we are done in this case. If a maximal path in G has 3 edges, then let these edges e (a, c, e 1 ), and pick e aritrarily from another component. Finally, if the maximal path in G has two edges, then let these edges e (a, c), and pick e 1 and e from two distinct components (this is possile since G is not the edge-disjoint union of two paths P ). Remark. Not only separating triangles impose constraints on the edge lengths. Consider the 4-connected triangulation T on the right, with a old path (a,, c, d, e, f). Note that Q = (, c, d, e) is a separating quadrilateral: In every emedding of T, either a or f lies in the interior of the polygon Q. In every emedding of T, the diameter of Q is less than l(c) + l(cd) + l(de). Hence, min{l(a), l(ef)} < l(c) + l(cd) + l(de), which is a nontrivial constraint for the edge lengths in G. e a d c f 3 Every Matching is Free In this section, we show that every matching M = (V, E) in every planar graph H is free. Given an aritrary length assignment for a matching M of H, we emed H with the specified edge lengths on M. Our algorithm is ased on a simple approach, which works well when M is well-separated (defined elow). In this case, we contract the edges in M to otain a triangulation Ĥ; emed Ĥ on a grid cz for a sufficiently large c > 0; and then expand the edges of M to the prescried lengths. If c > 0 is large enough, then the last step is only a small perturation of Ĥ, and we otain a valid emedding of H with prescried edge lengths. If, however, some edges in M appear in separating 3- or 4-cycles, then a significantly more involved machinery is necessary. 3.1 Edge Contraction and Vertex Splitting Operations A near-triangulation is a 3-connected planar graph in which all faces are triangles with at most one exception (which is typically considered to e the outer face). Let M e a matching in a planar graph H with a length assignment l: M R +. We may assume, y augmenting H if necessary, that H is a near-triangulation. Let D e an emedding of H where all the ounded faces are triangles. We shall construct a new emedding of H with the same vertices on the outer face where every edge e M has length l(e).

6 6 3 Every Matching is Free Edge contraction is an operation for a graph G = (V, E) and an edge e = v E: Delete and v and all incident edges; add a new vertex ˆv e ; and for every vertex u V \ {, v } adjacent to or v, add a new edge uˆv e. Suppose G is a near triangulation and v does not elong to a separating triangle. Then v is incident to at most two triangle faces, say v w 1 and v w, and so there are at most two vertices adjacent to oth and v. The cyclic sequence of neighors of ˆv e is composed of the sequence of neighors of from w 1 to w, and that of v from w to w 1 (in counterclockwise order). The inverse of an edge contraction is a vertex split operation that replaces a vertex ˆv e y an edge e = v. See Fig.. w w w v ˆv e ˆv e w1 w1 R e w 1 Figure : Left: An edge e = v of a near triangulation incident to the shaded triangles v w 1 and v w. Middle: e is contracted to a vertex ˆv e. The triangular faces incident to ˆv e form a star-shaped polygon. Right: We position edge e such that it contains ˆv e, and lies in the shaded doule wedge, and in the kernel of the star-shaped polygon centered at ˆv e. For simplicity, we consider only part of the doule wedge, lying in a rectangle R e of diameter ε. Suppose that we are given an emedding of a triangulation, and we would like to split an interior vertex ˆv e into an edge e = v such that (1) all other vertices remain at the same location; and () the common neighors of and v are w 1 and w (which are neighors of ˆv e ). Note that the ounded triangles incident to ˆv e form a star-shaped polygon, whose kernel contains ˆv e in the interior. We position e = v in the kernel of this star-shaped polygon such that the line segment e contains the point ˆv e, and vertices w 1 and w are on opposite sides of the supporting line of e. Therefore, e must lie in the doule wedge etween the supporting lines of ˆv e w 1 and ˆv e w (Fig., right). In Susection 3., we position e = v such that its midpoint is ˆv e ; and in Section 4, we place either or v at ˆv e, and place the other vertex in the appropriate wedge incident to ˆv e. 3. A Matching with Given Edge Lengths We now recursively prove that every matching in every planar graph is free. In one step of the recursion, we construct an emedding of a sugraph in the interior of a separating triangle (resp., a separating 4-cycle), where the length of one edge is given (resp., the lengths of two edges are given). The work done for a separating triangle or 4-cycle is summarized in the following lemma. Lemma 6 Let H = (V, E) e a near-triangulation and let M E e a matching with a length assignment l: M R +. (a) Suppose that a 3-cycle (, v, v 3 ), where v M, is a face of H. There is an L > 0 such that for every triangle ac with side length a = l( v ), c > L and ca > L, there is an emedding of H with prescried edge lengths where the outer face is ac and, v and v 3 are mapped to a, and c, respectively. () Suppose that a 4-cycle (, v, v 3, v 4 ), where v M and v 3 v 4 M, is a face of H. There is an L > 0 such that for every convex quadrilateral acd with side lengths a = l( v ), cd = l(v 3 v 4 ), ac > L, there is an emedding of H with prescried edge lengths where the outer face is acd and, v, v 3 and v 4 are mapped to a,, c and d, respectively.

7 Free Edge Lengths in Plane Graphs 7 Proof. We proceed y induction on the size of the matching M. We may assume, y applying an appropriate scaling, that min{l(e) : e M} = 1. (a) Consider an emedding D of H where v M is an edge of the outer face, and let M = M \ { v }. Let C 1,..., C k e the maximal separating triangles that include some edge from M, and the chordless separating 4-cycles that include two edges from M (more precisely, we consider all such separating triangles and separating chordless 4-cycles and among them we choose those that are not contained in the interior of any other such separating triangle or chordless 4-cycle). Let H 0 e the sugraph of H otained y deleting all vertices and incident edges lying in the interiors of the cycles C 1,..., C k. Let M 0 M denote the suset of edges of M contained in H 0. Let λ 0 = max{l(e) : e M 0 }. (1) For i = 1,..., k, let H i denote the sugraph of H that consists of the cycle C i and all vertices and edges that lie in C i in the emedding D; and let M i M e the suset of edges of M in H i. Applying induction for H i and M i, there is an L i > 0 such that H i can e emedded with the prescried lengths for the edges of M i in every triangle (resp., convex quadrilateral) with two edges of lengths at least L i. Let L = max{l i : i = 1,..., k}. By construction, M 0 is a well-separated matching in H 0 (recall that v is not in M 0 ). Successively contract every edge e = uv M 0 to a vertex ˆv e. We otain a planar graph Ĥ0 = ( V 0, Ê0) on at most n (and at least 3) vertices. Let D 0 e a grid emedding of Ĥ0 constructed y the algorithm of Kurowski [14], where the outer face is a triangle with vertices (0, 0), (3n 7, 0), and ( 3n 7, 3n 7 ); the only horizontal edge is the ase of the outer triangle; and the angular resolution of D 0 is ϱ 3 Ω(1/n). The minimum edge length is 5n 1, since all vertices have integer coordinates. There is an ε Ω(1/n) such that if we move each vertex of D 0 y at most ε, then the directions of the edges change y an angle less than ϱ/, and thus we retain an emedding. We could split each vertex ˆv e, e M, into an edge e that lies in the ε-disk centered at ˆv e, and in the doule wedge determined y the edges etween ˆv e and the common neighors of the endpoints of e (Fig., right). However, we shall split the vertices ˆv e, e M, only after applying the affine transformation α that maps the outer triangle of D 0 to a triangle ac such that α( ) = a, α(v ) = and α(v 3 ) = c. (The affine transformation α would distort the prescried edge lengths if we split the vertices now.) In the grid emedding D 0, the central angle of such a doule wedge is at least ϱ Ω(1/n), i.e., the angular resolution of D 0. The oundary of the doule wedge intersects the oundary of the ε-disk in four vertices of a rectangle that we denote y R e. Note that the center of R e is ˆv e, and its diameter is ε Ω(1/n). Hence, the aspect ratio of each R e, e M 0, is at least tan(ϱ/) Ω(1/n), and so the width of R e is Ω(1/n ). We show that if L = max{10n(l + λ 0 + a ), ξn 3 λ 0 }, for some constant ξ > 0, then the affine transformation α defined aove satisfies the following two conditions. The first condition allows splitting the vertices ˆv e, e M, into edges of desired lengths, and the second one ensures that the existing edges remain sufficiently long after the vertex splits. (i) every rectangle R e, e M 0, is mapped to a parallelogram α(r e ) of diameter at least λ 0 (defined in (1)); (ii) every nonhorizontal edge in D 0 is mapped to a segment of length at least L + λ 0. For (i), note that α maps a grid triangle of diameter 3n 7 < 3n into triangle ac of diameter more than L. Hence, it stretches every vector parallel to the preimage of the diameter of ac y a factor of at least L/(3n). Since the width of a rectangle R e, e M 0, is Ω(1/n ), the diameter of α(r e ) is at least Ω(L/n 3 ). If L Ω(n 3 λ 0 ) is sufficiently large, then the diameter of every α(r e ) is at least λ 0.

8 8 3 Every Matching is Free For (ii), we may assume w.l.o.g. that the triangle ac is positioned such that a = (0, 0) is the origin, = ( a, 0) is on the positive x-axis, and c is aove the x-axis (i.e., it has a positive y-coordinate). Then, the affine transformation α is a linear transformation with an upper triangular matrix: α ([ x y ]) = [ A B 0 C ] [ x y ] = [ Ax + By Cy where A, C > 0, and y symmetry we may assume B 0. We show that if L 10n(L + λ 0 + a ), then α maps every nonhorizontal edge of D 0 to a segment of length at least L + λ 0. A nonhorizontal edge in the grid emedding D 0, directed upward, is an integer vector (x, y) with x [ 3n + 7, 3n 7] and y [1, 3n 7 ]. It is enough to show that (Ax + By) + (Cy) > (L + λ 0 ) for x [ 3n, 3n] and y [1, 3 n]. Since α maps the right corner of the outer grid triangle (3n 7, 0) to = ( a, 0), we have A = a /(3n 7). Since ac > L, where a = (0, 0) and c = α ( ( 3n 7, 3n 7 ) ), we have ( A 3n 7 + B 3n 7 ) ( + C 3n 7 ) = ac > L 100n (L + λ 0 + a ). () We distinguish two cases ased on which term is dominant in the left hand side of (): Case 1: (C 3n 7 ) 50n (L + λ 0 + a ). In this case, we have C > (L + λ 0 ), and so (Cy) > (L + λ 0 ) since y 1. Case : (A 3n 7 +B 3n 7 ) > 50n (L +λ 0 + a ). In this case, we have A 3n 7 +B 3n 7 > 7n(L + λ 0 + a ). Comined with A = a /(3n 7), this gives B > 4(L + λ 0 + a ). It follows that (Ax + By) > (L + λ 0 ), as claimed, since Ax a and y 1. We can now reverse the edge contraction operations, that is, split each vertex ˆv e, e M 0, into an edge e of length l(e) within the parallelogram α(r e ). By (i), we otain an emedding of H 0. Each cycle C i, i = 1,..., k, is a triangle (resp., quadrilateral) where the edges of M 0 have prescried lengths, and any other edge has length at least L = max{l i : i = 1,..., k} y (ii). By induction, we can insert an emedding of H i with prescried lengths on the matching M i into the emedding of the cycle C i, for i = 1,..., k. We otain the required emedding of H. () The proof for the case when the outer face of H is a 4-cycle follows the same strategy as for (a), with some additional twists. Suppose we are given a convex quadrilateral acd as descried in the statement of the lemma. Denote y q the intersection of its diagonals. We show that ( aq and q are oth at least L/3) or ( cq and dq are oth at least L/3) if L > 9 max( a, cd ). Indeed, we have ac > c a > 8 9L from the triangle inequality for ac. Since ac = aq + cq, we have aq > 4 9 L or cq > 4 9 L. If aq > 4 9L, then q > aq a > 1 3 L from the triangle inequality for aq; otherwise dq > cq cd > 1 3L. We may assume without loss of generality that aq > L/3 and q > L/3. In the remainder of the proof, we emed H such that almost all vertices lie in the triangle aq, and the vertices, v, v 3, and v 4 are mapped to a,, c, and d, respectively. Similarly to (a), we define H 0 as the graph otained y deleting all vertices and incident edges lying in the interior of maximal separating triangles or chordless 4-cycles, containing an edge from M \ { v }. Define L as efore, y using the inductive hypothesis in the separating cycles. Contract successively all remaining edges of M \ { v } that are in H 0 (including edge v 3 v 4 ) to otain a graph Ĥ0. Denote y ˆv 3 the vertex of Ĥ 0 corresponding to v 3 v 4 M, and consider an emedding of Ĥ0 with the outer face v ˆv 3. We again use the emedding D 0 of Kurowski [14], such that, v and ˆv 3 are mapped to (0, 0), (3n 7, 0), and ( 3n 7, 3n 7 ), respectively. We first split vertex ˆv 3 into two vertices v 3 and v 4, exploiting the fact ],

9 Free Edge Lengths in Plane Graphs 9 v 3 v 3 v 4 p v 4 p ˆv 3 ˆv 3 u 1 = u 0 u k = v u 1 = u 0 u k = v Figure 3: Left: The emedding D 0 into a triangle v ˆv 3, and the x-monotone path = u 0, u 1,..., u k formed y the neighors of ˆV 3. A point p lies aove ˆv 3, and the rays emitted y p in directions (1, ) and ( 1, ). Right: Vertex ˆv 3 is split into v 3 and v 4 on the two rays emitted y p. that ˆv 3 is a oundary vertex in D 0 and some special properties of the emedding in [14] (descried elow); and then split all other contracted vertices of Ĥ0 similarly to (a). Denote the neighors of ˆv 3 in D 0 in counterclockwise order y = u 0, u 1,..., u k = v (Fig. 3, left). The grid emedding in [14] has the following property (mentioned in Section 1): the path u 0,..., u k is x-monotone and the slope of every edge is in the range ( 1, 1). Let p = ( 3n 7, 3n 7 ), and note that the slope of every line etween p and u 1,..., u k, is outside of the range (, ). Similarly, if we place the points v 3 (resp., v 4 ) on the ray emitted y p in direction (1, ) (resp., ( 1, )), then the slope of every line etween v 3 (resp., v 4 ) and u 1,..., u k is outside of (, ). We can now split vertex ˆv 3 as follows. Refer to Fig. 3. Let α e the affine transformation that maps the triangle v p to aq such that α( ) = a, α(v ) = and α(p) = q. Since the diagonals ac and d intersect at q, the segments α 1 (c) and v α 1 (d) intersect at p. We split vertex ˆv 3 into v 3 = α 1 (c) and v 4 = α 1 (d). By the aove oservation, the edges incident to v 3 and v 4 remain aove the x-monotone path u 0,..., u k. (Note, however, that the angles etween edges incident with v 3 or v 4 may e aritrarily small.) With a very similar computation as for (a), we conclude that for a large enough L Ω(L + λ 0 ) we can guarantee the same two properties we needed in (a), that is, α maps every small rectangle R e to a parallelogram α(r e ) whose diameter is at least λ 0, and every nonhorizontal edge to a segment of length at least L + λ 0. Hence, every remaining contracted vertex v e in D 0 can e split within the parallelogram α(r e ) as in (a). To finish the construction, it remains to apply the inductive hypothesis to fill in the missing parts in the maximal separating triangles or 4-cycles. We are now ready to prove the main result of this section. Theorem 7 Every matching in a planar graph is free. Proof. Let H = (V, E) e a planar graph, and let M E e a matching with a length assignment l: M R +. We may assume, y augmenting H with new edges if necessary, that H is a triangulation. Consider an emedding of H such that an edge e M is on the outer face. Now Lemma 6 completes the proof. 4 Graphs with Three or Four Edges By Theorems 5 and 7, a graph G with at least five edges is free in every host H if and only if G is a matching. For graphs with four edges, the situation is also clear except for the case of the disjoint union of two paths

10 10 4 Graphs with Three or Four Edges of two edges each. In this section we show that every forest with three edges, as well as the disjoint union of two paths of length two are always free. We show (Lemma 9) that it is enough to consider hosts H in which G is a spanning sugraph, that is, V (G) = V (H). For a planar graph G = (V, E), the triangulation of G is an edge-maximal planar graph T, G T, on the vertex set V. (The following lemma holds for every graph G, including matchings. However, it would not simplify the argument in that case.) Lemma 8 If G is a sugraph of a triangulation H with 0 < V (G) < V (H), then there is an edge in H etween a vertex in V (H) and a vertex in V (H) V (G) that does not elong to any separating triangle of H. Proof. Let V = V (G) denote the vertex set of G and U = V (H) \ V. Let E(U, V ) e the set of edges in H etween U and V. Since H is connected, E(U, V ) is nonempty. Consider an aritrary emedding of H (with aritrary edge lengths). For every edge uv E(U, V ), let k(uv) denote the maximum numer of vertices of H that lie in the interior of a triangle (u, v, w) of H, where w V (H). Let uv E(U, V ) e an edge that minimizes k(uv). If k(uv) = 0, then uv does not elong to any separating triangle, as claimed. For the sake of contradiction, suppose k(uv) > 0, and let (u, v, w) e a triangle in H that contains exactly k(uv) vertices of H. Since H is a triangulation, there is a path etween u and v via the interior of (u, v, w). Since u U and v V, one edge of this path must e in E(U, V ), say u v E(U, V ). Note that any triangle (u, v, w ) of H lies inside the triangle (u, v, w), and hence contains strictly fewer vertices than (u, v, w). Hence k(u v ) < k(u, v) contradicting the choice of edge uv. Lemma 9 If a planar graph G is (extrinsically) free in every triangulation of G, then G is (extrinsically) free in every planar host H, G H. Proof. Let G = (V, E) e a planar graph with a length assignment l : E R +. It is enough to prove that G is (extrinsically) free in every triangulation H, G H. We proceed y induction on n = V (H) V (G), the numer of extra vertices in the host H. If n = 0, then H is a triangulation of G, and G is free in H y assumption. Consider a triangulation H, G H, and assume that the claim holds for all smaller triangulations H, G H. By Lemma 8, there is an edge e = uv in H etween v V (G) and u V (H) V (G) that does not elong to any separating triangle. Contract e into a vertex ˆv e to otain a triangulation H, G H. By induction, H admits a straight-line emedding in which the edges of G have prescried lengths. Since e is not part of a separating triangle of H, we can split vertex ˆv e into u and v such that v is located at point ˆv e, and u lies in a sufficiently small neighorhood of ˆv e (refer to Fig.). Thus, we otained a straight line emedding of H in which edges of G have prescried lengths. The next theorem finishes the characterization of free graphs. Theorem 10 Let G e a sugraph of a planar graph H, such that G is (1) the star with three edges; or () the path with three edges; or (3) the disjoint union of a path with two edges and a path with one edge; or (4) the disjoint union of two paths with two edges each. Then G is free in H. Proof. By Lemma 9 it is enough to prove the theorem in the case when G is a spanning sugraph of H. We can also assume that H is a triangulation.

11 Free Edge Lengths in Plane Graphs 11 (1) If G is the star with three edges, then H is K 4. Emed the center of the star at the origin. Place the three leaves on three rotationally symmetric rays emitted y the origin, at prescried lengths from the origin (Fig. 4(a)). The remaining three edges are emedded as straight line segments on the convex hull of the three leaves. () Let G e the path (, v, v 3, v 4 ) with l( v ) l(v 3 v 4 ). Emed v at the origin, place and v 3 on the positive x- and y-axis respectively, at prescried distance from v. Note that = conv(, v, v 3 ) is a right triangle whose diameter (hypotenuse) is larger than the other two sides (Fig. 4(a)). Thus we can emed v 4 at a point in the interior of at distance l(v 3 v 4 ) from v 3. Since the four vertices have a triangular convex hull, H = K 4 emeds as a straight-line graph. (3) Suppose that G = (V, E) is the disjoint union of path (, v, v 3 ) and (v 4, v 5 ). Since H has five vertices there exists at most one separating triangle in H. Thus, the path of G with two edges contains an edge, say e = v, that does not elong to any separating triangle. Contract edge e to a vertex ˆv e, otaining a triangulation H = K 4 on four vertices, and a perfect matching G H. Let us emed the two edges of G with prescried lengths such that one lies on the x-axis, the other lies on the orthogonal isector of the first edge at distance l(e) from the x-axis. This defines a straight-line emedding of H, as well. We otain a desired emedding of H y splitting vertex ˆv e into edge e such that v is emedded at point ˆv e and is mapped to a point in the kernel of the appropriate star-shaped polygon (c.f. Fig. ). By the choice of our emedding of H, the diameter of this kernel is more than l(e), and we can split ˆv e without introducing any edge crossing (Fig.4(c)) v 3 v 3 v 3 ˆv e v v 4 (a) v () v 4 v 4 v 5 v5 ˆv e (c) (d) ˆv e Figure 4: (a) Emedding of a star with three leaves. () Emedding of a path of three edges. (c) Graph H and regions that are used for splitting ˆv e. (d) Graph H and its spanning sugraph G whose edges have prescried lengths. (4) Assume that G is the disjoint union of two paths P 1 and P, each with two edges. Since G is a spanning sugraph of H, neither path can span a separating triangle. Moreover, since there exist at most two separating triangles in H, one of the paths, say P 1, contains an edge e that is not part of a separating triangle. Contract edge e to ˆv e, otaining a triangulation H and a sugraph G. Similarly to the case (3), emed H respecting the lengths of all the edges of G such that all edges etween the two components of G have length at least l(e). By the choice of our drawing of H, the kernel of the appropriate star-shaped polygon has diameter at least l(e). Therefore, we can split e into two vertices such that the middle vertex of P 1 remains at ˆv e, and the endpoint of P 1 is emedded at distance l(e) from ˆv e (Fig. 4(d)). 5 Stars are Free in 4-Connected Triangulations In this section, we prove Theorem. Theorem Every star is free in a 4-connected triangulation. Proof. Let T = (V, E) e a 4-connected triangulation. Let S E e a star centered at v 0 V with k 3 edges S = {v 0,..., v 0 v k } laeled cyclically around v 0. Let l: S R + e a length assignment. We emed T such that every edge e S has length l(e). Refer to Fig. 5(upper-left).

12 1 5 Stars are Free in 4-Connected Triangulations Since T is a triangulation, every two consecutive neighors of v 0 are adjacent. Nonconsecutive neighors of v 0 are nonadjacent, otherwise they would create a separating triangle. In other words, the vertices {,..., v k } induce a cycle C = (,..., v k ) in T. Consider an emedding D of T in which v 0 is an interior vertex. Note that the outer face is a triangle, since T is a triangulation. If two neighors of v 0 are incident on the outer triangle, they must e consecutive neighors of v 0, since T is 4-connected. Therefore, if v 0 has 3 neighors on the outer face, then T = K 4, and it oviously has an emedding with prescried edge lengths. We can distinguish two cases: Case 1: at most one neighor of v 0 is incident to the outer face in D. Refer to Fig. 5. Then every edge of C is an interior edge of D. The two triangles adjacent to every edge v i v i+1 form a quadrilateral incident to v 0 and some other vertex, which is nonadjacent to v 0 (otherwise there would e a separating triangle in T ). Consider all 4-cycles incident to v 0 and to a vertex nonadjacent to v 0. Such a cycle is called maximal if it is not contained in any other such 4-cycle. Let C 1, C,..., C m e a collection of maximal 4-cycles, each incident to v 0 and to some nonadjacent vertex, in counterclockwise order around v 0. Denote y u i, i = 1,..., m, the vertex in C i that is not adjacent to v 0. It is clear that every triangle incident to v 0 is contained in one of the cycles C i, every cycle C i passes through exactly two edges of the star H, and the numer of cycles is at least m. For i = 1,..., m, the consecutive cycles C i and C i+1 (with m + 1 = 1) share exactly one edge, y their maximality, which is denoted v 0 v κ(i). u 1 u 1 v 7 v 7 C 1 v C 0 u v 4 11 u 3 u 4 u C C 4 3 v 4 11 u 3 u 4 u 1 u 1 ˆv 0 v 7 u 1 4 u u 3 u 3 u 4 Figure 5: Upper-left: A star centered at v 0 in a 4-connected triangulation. The maximal 4-cycles C 1,..., C 4 incident to v 0 and to some nonadjacent vertex are defined y vertices u 1,..., u 4. Upper-right: The graph otained after deleting all vertices and incident edges in the interior of the cycles C 1,..., C 4. Lower-left: The remaining edges of the star are contracted to ˆv 0. The cycles C 1,... C 4 collapse to old edges. The faces incident to ˆv 0 form a star-shaped polygon, whose kernel contains an ε-neighorhood of ˆv 0. Lower-Right: v 0 is emedded at ˆv 0, and vertices v κ(j) of cycles C j are emedded on the segment ˆv 0 u j for j = 1,..., 4. u 4

13 Free Edge Lengths in Plane Graphs 13 We construct a triangulation T = ( V, Ê) in two steps: First delete all vertices and incident edges in the interiors of the cycles C 1,..., C m (Fig. 5, upper-right); and then successively contract the remaining edges v 0 v κ(1),..., v 0 v κ(m) of H (Fig. 5, lower-left). The vertices v 0,,..., v k merge into a single vertex ˆv 0 in T, and the cycles C i collapse into distinct edges incident to ˆv 0. Consider an emedding D of T in which ˆv 0 is an interior vertex. The triangles incident to ˆv 0 form a star-shaped polygon P, whose vertices include u 1,..., u m, and vertex ˆv 0 lies in the interior of the kernel of P. There is an ε > 0 such that the ε-neighorhood of ˆv 0 also lies in the kernel of P. We may assume (y scaling) that ε = max{l(e) : e M}. v 7 v 0 u 1 v 6 v 5 v 4 v 0 v 7 v 3 v c v v 6 5 v 4 v 3 v Figure 6: Left: vertices v i, i =,..., 6, are successively emedded in the triangles v 0 v i 1 u 1. Right: triangle v 0 v is emedded such that v 0 and v 0 v are almost collinear; and then v i, i = 3,..., k, are successively emedded in the triangles v 0 v i 1 c. Emed the star center v 0 at ˆv 0 ; and for j = 1,..., m, emed vertex v κ(j) at the point on the segment v 0 u j at distance l(v 0 v κ(j) ) from the center (Fig. 5, lower-right). Hence each cycle C j = (v 0, v κ(j 1), u j, v κ(j) ) is emedded in (weakly) convex position, where edges v 0 v κ(j) and v κ(j) u j are collinear. For j = 1,..., m, the vertices v κ(j)+1,..., v κ(j+1) 1 should lie in the interior of the triangle j = v 0 u j v κ(j+1). Emed successively v κ(j)+1,..., v κ(j+1) 1 at the prescried distance from the center v 0 in the sequence of nested triangles v 0 u j v i 1 (Fig. 6, left). As a result, the path (v κ(j),..., v κ(j+1) ) partitions the triangle j into two star-shaped polygons, with star centers v 0 and u j, respectively. All remaining vertices in the interior of C j can e emedded in the latter star-shaped polygon y the Hong-Nagamochi theorem [11]. Case : exactly two neighors of v 0 are on the outer triangle. Without loss of generality the outer face of D is the triangle v c. Refer to Fig. 6(right). Emed triangle v 0 v such that v 0 and v 0 v have prescried edge lengths, and v has length l(v 0 ) + l(v 0 v ) ε for a small ε > 0. Emed vertex c at distance max{l(v 0 v i ) : 1 i k} from v 0 such that the line v 0 c is orthogonal to v. Emed successively v 3,..., v k in the sequence of nested triangles v 0 v i 1 c. Then the path (v, v 3..., v m, ) partitions the triangle v c into two star-shaped polygons, with star centers v 0 and c, respectively. All remaining vertices in the interior of T can e emedded in the latter star-shaped polygon y the Hong- Nagamochi theorem [11]. 6 Extrinsically Free Sugraphs In this section, we prove Theorem 3. A forest G = (V, E) has a straight-line emedding with every length assignment l: E R +, that is, it has no intrinsic constraints on the edge lengths. Theorems 5 and 7 classify all forests G that are extrinsically free in every host H. It remains to classify all planar graphs G that contain cycles, except for the cases that G itself is a cycle with k 4 vertices. Our positive results are limited to two types of graphs.

14 14 6 Extrinsically Free Sugraphs Lemma 11 A sugraph G of a planar graph H is extrinsically free if G is (1) a triangulation; or () a triangle and one additional edge (either disjoint from or incident to the triangle). Proof. We may assume that H is a triangulation, y augmenting H with dummy edges if necessary. (1) Let G = (V, E) e a triangulation with a length assignment l: E R +, such that G has an emedding D 0 with edge length l(e), e E; and let D 1 e an aritrary emedding of H. Since every triangulation has a unique cominatorial emedding, D 0 is cominatorially equivalent to the restriction of D 1 to G. Partition the vertices into the vertex sets of the connected components of H \ G, each lying in a face of G. By Tutte s arycenter method [0], we can emed each vertex class within the corresponding triangular face of G 0. () Let ac and e e a triangle and an edge in G. Consider an emedding of H such that a is an edge of the outer face and e lies outside of ac. An argument analogous to Lemma 6(1) shows that H has an emedding such that the edges of the triangle ac are mapped to a given triangle and in the exterior of that triangle edge e has prescried length. (Recall that in Lemma 6(1), the outer face was mapped to a given triangle such that disjoint edges in the interior of the triangle had prescried lengths.) In the remainder of this section, we show that no other planar graph G is extrinsically free in all hosts H if G contains a cycle. We start y oserving that it is enough to consider graphs with at most two components. Oservation 3 Let G = (V, E) e extrinsically free in every host H, G H. If G contains a cycle, then G has at most two connected components. Proof. Suppose to the contrary that G has at least three connected components, denoted G 1, G, and G 3. Assume, without loss of generality, that G 1 contains a cycle C. Augment G to a triangulation in which C is a separating cycle, separating G and G 3. Let l: E R + e a length assignment that assigns a total length of 1 to the edges of C, and length at least 1 to every edge of G and G 3, respectively. In every emedding of H, one of G and G 3 lies in the interior of the simple polygon C, however, every edge in G and G 3 is longer than the diameter of C. Hence H cannot e emedded with the prescried edge lengths, and so G is not extrinsically free. The following technical lemma is the key tool for treating the remaining cases except for when G is a cycle. Lemma 1 Let G = (V, E) e a planar graph such that two vertices a, c V are connected y three independent paths P 1, P, and P 3 ; the paths P and P 3 have some interior vertices; and the interior vertices of P and P 3 are in distinct components of G P 1. Then G is not extrinsically free in some host H, G H. Proof. We may assume that H is a triangulation, y augmenting H with dummy edges if necessary. Furthermore, we may also assume that P 1, P and P 3 are three independent paths with the aove properties that have the minimum total numer of vertices in G. We start with a rief overview of the proof: We shall descrie an emedding D 1 of G; then define H = (V, E {uv}) y augmenting G with a single new edge uv (to e determined); and finally show that no emedding of G with the same edge lengths as in D 1 can e augmented to an emedding of H. Refer to Fig. 7(left). Denote y ac 1 the edge of P 1 incident to a, with possily c 1 = c. Denote y a 1 and c the edges of P incident to its endpoints (possily 1 = ); and y ad 1 and cd the edges of P 3 incident to its endpoints (possily d 1 = d ). Consider an emedding D 0 of G such that P 1 lies inside the cycle P P 3. The vertices of G that are incident to none of the three paths P 1, P, P 3, can e partitioned as follows: Let V and V d denote the vertices lying in the interior of cycles P 1 P and P 1 P 3, respectively. The vertices in the exterior of

15 Free Edge Lengths in Plane Graphs 15 G c 1 G P 1 P 1 a P 1 c 1 c P c 1 P 3 a P 3 d 1 = d d 1 = d Figure 7: Left: emedding D 0 of G, where path P 1 separates P from P 3. Right: emedding D 1 of G where path P 1 lies on the x-axis, and P P 3 lies in an ε-neighorhood of the parallelogram a 1 cd 1. cycle P P 3 can e partitioned into two sets, since G contains no path etween interior vertices of P and P 3 : Let V + and V + d denote the vertices lying in the exterior of cycle P P 3 and joined with P or P 3, respectively, y a path in G P 1. We now construct a straight-line emedding D 1 cominatorially equivalent to D 0 as follows (refer to Fig. 7, right). Let 0 < ε < 1/( E ) e a small constant. Emed path P 1 on the positive x-axis with a = (0, 0), c = (3, 0), and if c 1 c, then c 1 = (0, 3 ε). Let 1 = (1, 1), and if 1, then emed the vertices of P etween 1 and on a horizontal segment etween 1 = (1, 1) and = (1+ε, 1). Similarly, let d 1 = (, 1), and if d 1 d, then emed the vertices of P 3 etween d 1 and d on a horizontal segment etween d 1 = (, 1) and d = ( + ε, 1). The vertices in V (resp., V d ) can e emedded y Tutte s arycenter method [0], since the cycle P 1 P (resp., P 1 P 3 ) is emedded as a (weakly) convex polygon, and the path P (resp., P 3 ) has no shortcut edges y the choice of P 1, P and P 3. The vertices of V + (resp., V + d ) can e emedded y the Hong-Nagamochi theorem in an ε-neighorhood of 1 (resp., d 1 ), since the region aove P (resp., elow P 3 ) is star-shaped. This completes the description of D 1. Let H = (V, E {uv}) where u (resp., v) is a vertex of the outer face strictly aove (resp., elow) the x-axis in the emedding D 1. Note that vertices u and v are in an ε-neighorhood of 1 and d 1 respectively, and so cannot e connected y a straight edge in D 1. Let D e an emedding of G in which every edge has the same length as in D 1. Assume, without loss of generality, that a = (0, 0) and ac 1 lies on the positive x-axis, and 1 is aove the x-axis. By the length constraints, path P 1 lies in the ε-neighorhood of edge ac 1. Paths P lies in the ε-neighorhood of its position in D 1. Vertex d 1 must e elow the x-axis, otherwise c and ad 1 would cross. Hence path P 3 is also in the ε-neighorhood of its location in D 1. All interior vertices of P and all vertices in V + are in the (ε E )-neighorhood of 1. Similarly, all interior vertices of P 3 and all vertices in V + d are in the (ε E )- neighorhood of d 1. Since ε E < 1, the line segment uv crosses ac 1, and so H cannot e emedded with the prescried edge lengths. This confirms that G is not extrinsically free in H. Lemma 13 Let G = (V, E) e extrinsically free in every host H, G H. If G contains a cycle C with k 4 vertices such that all vertices of C are incident to a common face in some emedding of G, then C is a -connected component of G. Proof. Consider an emedding D 0 of G in which all vertices of C are incident to common face F. Assume without loss of generality that F is a ounded face that lies inside C. We first show that C must e a chordless cycle. Indeed, if C has an (exterior) chord ac, then G would not e extrinsically free y applying Lemma 1 with the three paths P 1 = {ac}, and letting P and P 3 e the two arcs of C etween a and c. Let a and c e two nonadjacent vertices of C (Fig. 8). If there is a path P etween a and c (via the interior or exterior of C), then again G would not e extrinsically free y applying Lemma 1 for P and the two arcs