Pentagon contact representations

Transcription

1 Pentagon contact representations Stean Felsner Institut ür Mathematik Technische Universität Berlin Hendrik Schrezenmaier Institut ür Mathematik Technische Universität Berlin August, 017 Raphael Steiner Fakultät ür Mathematik und Inormatik FernUniversität in Hagen Abstract. Representations o planar triangulations as contact graphs o a set o internally disjoint homothetic triangles or respectively o a set o internally disjoint homothetic squares have received quite some attention in recent years. In this paper we investigate representations o planar triangulations as contact graphs o a set o internally disjoint homothetic pentagons. Surprisingly such a representation exists or every triangulation whose outer ace is a 5-gon. We relate these representations to ive color orests. These combinatorial structures resemble Schnyder woods respectively transversal structures. In particular there is a bijection to certain α-orientations and consequently a lattice structure on the set o ive color orests o a given graph. This lattice structure plays a role in an algorithm that is supposed to compute a contact representation with pentagons or a given graph. Based on a ive color orest the algorithm builds a system o linear equations and solves it, i the solution is non-negative, it encodes distances between corners o a pentagon representation. In this case the representation is constructed and the algorithm terminates. Otherwise negative variables guide a change o the ive color orest and the procedure is restarted with the new ive color orest. Similar algorithms have been proposed or contact representations with homothetic triangles and with squares. 1 Introduction A pentagon contact system S is a inite system o convex pentagons in the plane such that any two pentagons intersect in at most one point. I all pentagons o S are regular pentagons with a horizontal side at the bottom, we call S a regular pentagon contact representation. Note that in this case any two pentagons o S are homothetic. The contact system is non-degenerate i every contact involves exactly one corner o a pentagon. The contact graph G (S) o S is the graph that has a vertex or every pentagon and an edge or every contact o two pentagons in S. Note that G (S) inherits a crossing-ree embedding into the plane rom S. For a given plane graph G and a pentagon contact system S with G (S) = G we say that S is a pentagon contact representation o G. We will only consider the case that G is an inner triangulation o a 5-gon, i.e., the outer ace o G is a 5-gon with vertices a 1,..., a 5 in clockwise order, all inner aces are triangles, there The conerence version o this paper has appeared in the proceedings o Eurocomb 17 (ENDM 61C, pp ) under the title Pentagon contact representations. Partially supported by DFG grant FE-340/

2 a 1 a 5 a a 4 a 3 Figure 1: A regular pentagon contact representation o the graph shown in black. are no loops or multiple edges, and the only edges between the vertices a 1,..., a 5 are the ive edges o the outer ace. Our interest lies in regular pentagon contact representations o G with the additional property that a 1,..., a 5 are represented by line segments s 1,..., s 5 which together orm a pentagon with all internal angles equal to (3/5)π. The line segment s 1 is always horizontal and at the top, and s 1,..., s 5 is the clockwise order o the segments o the pentagon. Fig. 1 shows an example. Triangle contact representations have been introduced by De Fraysseix et al. [5]. They observed that Schnyder woods can be considered as combinatorial encodings o triangle contact representations o triangulations and essentially showed that any Schnyder wood can be used to construct a corresponding triangle contact system. They also showed that the triangles can be requested to be isosceles with a horizontal basis. Representations with homothetic triangles can degenerate in the presence o separating triangles. Gonçalves et al. [9] showed that 4-connected triangulations admit contact representations with homothetic triangles. The proo is an application o Schramm s Convex Packing Theorem, a strong theorem which is based on his Monster Packing Theorem. A more combinatorial approach to homothetic triangle contact representations which aims at computing the representation as the solution o a system o linear equations related to a Schnyder wood was described by Felsner [7]. On the basis o this approach Schrezenmaier reproved the existence o homothetic triangle representations in his Masters thesis [15]. Representations o graphs using squares or more precisely graphs as a tool to model packings o squares already appear in classical work o Brooks et al. [3] rom Schramm [14] proved that every 5-connected inner triangulation o a 4-gon admits a square contact representation. Again there is a combinatorial approach to this result which aims at computing the representation as the solution o a system o linear equations, see Felsner [8]. In this instance the role o Schnyder woods is taken by transversal structures. As in the case o homothetic triangles this approach comes with an algorithm which works well in practice, however, the proo that the algorithm terminates with a solution is still missing. On the basis o the non-algorithmic aspects o this approach Schrezenmaier [15] reproved Schramm s Squaring Theorem. In this paper we investigate representations o planar triangulations as contact graphs o a set o internally disjoint homothetic pentagons. From Schramm s Convex Packing Theorem it easily ollows that such a representation exists or every triangulation whose outer ace is a 5-gon. We relate such representations to ive color orests. The main part o the paper is devoted to the study o this combinatorial structure. It will become clear that ive color orests are close relatives o Schnyder woods and transversal structures. We note in passing that Bernardi and Fusy [1] also studied some relatives o Schnyder woods using 5 colors. Their ive color trees, however, only

3 P a1 P a5 P a P a4 P a3 Figure : Prototypes or the ive outer vertices o G. live on duals o 5-regular planar graphs. At the end o the paper we propose an algorithm or computing homothetic pentagon representations on the basis o systems o equations and local changes in the corresponding ive color orests. We conjecture that the algorithm always terminates. A proo o this conjecture would imply a proo or the existence o pentagon contact representations which is independent o Schramm s Monster Packing. The idea o looking or pentagon contact representations and a substantial part o the work originate in the Bachelors Thesis o Steiner [16]. 1.1 The existence o pentagon contact representations The existence o regular pentagon contact representations or every inner triangulation o a 5-gon can be shown using the ollowing general result about contact representations by Schramm. Theorem 1 (Convex Packing Theorem [13]) Let G be an inner triangulation o the triangle abc. Further let C be a simple closed curve in the plane partitioned into three arcs P a, P b, P c, and or each interior vertex v o G let P v be a convex set in the plane containing more than one point. Then there exists a contact representation o a supergraph o G (on the same vertex set, but possibly with more edges) where each interior vertex v is represented by a single point or a homothetic copy o its prototype P v and each outer vertex w by the arc P w. Theorem Let G be an inner triangulation o the 5-gon a 1,..., a 5. Then there exists a regular pentagon contact representation o G. Proo. By adding the edges a 1 a 3 and a 1 a 4 in the outer ace o G, it becomes a triangulation G with outer ace a 1 a 3 a 4. We deine the arcs P a1, P a3, P a4 to be extensions o the upper, lower let, respectively lower right edge o a regular pentagon A with a horizontal edge at the top, such that P a1 P a3 P a4 orm a triangle and thereore a simple closed curve. We deine the convex sets P a and P a5 to be line segments parallel to the upper right and upper let edge o the pentagon A (see Fig. ). Finally, or each interior vertex v o G let P v be a regular pentagon with a horizontal edge at the bottom. Now we can apply the Convex Packing Theorem. The result is a contact representation o a supergraph o G where a 1, a 3, a 4 are represented by P a1, P a3, P a4 and every inner vertex v by a homothetic copy o P v or by a single point. We claim that in this contact representation o G none o the homothetic copies o the prototypes is degenerate to a single point. Assume there is a degenerate copy in the contact representation. Let H be a maximal connected component o the subgraph o G induced by the vertices whose pentagons are degenerate to a single point. Since the line segments corresponding 3

4 a 1 a a a 4 a 3 Figure 3: The local conditions o a ive color orest to the three outer vertices are not degenerate, H has to be bounded by a cycle C o vertices whose pentagons respectively line segments are not degenerate. In the contact representation all vertices o H are represented by the same point and thereore all pentagons respectively line segments representing the vertices o C have a contact with this point, i.e., they meet at the point. But or geometric reasons at most two o these can meet in a single point. Thus C is a -cycle, in contradiction to our deinition o inner triangulations that does not allow multiple edges. Ater cutting the segments P a1, P a3 and P a4, the vertices a 1,..., a 5 are represented by a pentagon o the required orm and we obtain a regular pentagon contact representation o G. Five Color Forests In this section G will always be a triangulation with outer ace a 1,..., a 5 in clockwise order. The set 1,..., 5 o colors is to be understood as representatives modulo 5, e.g., 1 and 4 denote the same color. Deinition 1. A ive color orest o G is an orientation and coloring o the inner edges o G in the colors 1,..., 5 with the ollowing properties (see Fig. 3 or an illustration): (F1) All edges incident to a i are oriented towards a i and colored in the color i. (F) For each inner vertex v the incoming edges build ive (possibly empty) blocks B i, i = 1,..., 5, o edges o color i and the clockwise order o these blocks is B 1,..., B 5. Moreover v has at most one outgoing edge o color i and such an edge has to be located between the blocks B i+ and B i. (F3) For every inner vertex and or i = 1,..., 5 the block B i is nonempty or one o the outgoing edges o colors i and i + exists. The ollowing theorem shows the key correspondence between ive color orests and pentagon contact representations. Theorem 3 Every regular pentagon contact representation induces a ive color orest on its contact graph. Proo. Let S be a regular pentagon contact representation o G = G (S). We color the corners o all pentagons o S with the colors 1,..., 5 in clockwise order, starting with color 1 at the corner opposite to the horizontal segment. Let e be an inner edge o G. I e corresponds to the contact o a corner o a pentagon A and a side o a pentagon B in S, then we orient the edge e 4

5 Figure 4: The induced ive color orest and α-orientation o a pentagon contact representation. rom the vertex corresponding to A to the vertex corresponding to B and color it in the color o the corner o A involved in the contact. A contact o two pentagon corners can be interpreted in two ways as a corner-side contact with ininitesimal distance to the other corner. We choose one o these interpretations and proceed as beore. Hence, the ive color orest induced by a degenerate pentagon contact representation is not unique. Figure 4 shows an example. We claim that this coloring and orientation o G ulills the properties o a ive color orest. Property (F1) immediately ollows rom the construction. Now consider property (F). It is clear that every inner vertex has at most one outgoing edge o every color. That the incoming edges lie in the right interval, ollows rom the act that a corner-side contact between two homothetic regular pentagons is only possible i a corner o the irst pentagon touches the opposite side o the second pentagon. Finally we check property (F3) or the case i = 1. The other cases are symmetric. Let v be an inner vertex o G and A the corresponding pentagon o S. Since G is a triangulation, the pentagons corresponding to any two consecutive neighbors o v have to touch. Thereore a pentagon B o these pentagons has to intersect the area below the horizontal side o A, and that is only possible i the contact o A and B corresponds to a an incoming edge o v o color 1 or an outgoing edge o color 3 or 4. Schnyder [1] introduced a similar structure or inner triangulations o a triangle: Deinition. A Schnyder wood o an inner triangulation T o the triangle b 1, b, b 3 is an orientation and coloring o the inner edges o T in the colors 1,, 3 with the ollowing properties: (S1) All edges incident to b i are oriented towards b i and colored in the color i. (S) Each inner vertex has in clockwise order exactly one outgoing edge o color 1, one outgoing edge o color and one outgoing edge o color 3, and in the interval between two outgoing edges there are only incoming edges in the third color. Schnyder proved that every inner triangulation o a triangle admits a Schnyder wood, and using this result, we will show that every inner triangulation o a pentagon admits a ive color orest. Theorem 4 ([1]) Let T be an inner triangulation o a triangle. Then there exists a Schnyder wood o T. Theorem 5 Let G be an inner triangulation o the pentagon a 1,..., a 5. Then there exists a ive color orest o G. 5

6 a 1 b 1 b 1 a 1 a 5 a a 5 a b 4 c c 5 b 3 c c 5 c c 5 a 4 a 3 b 4 b 3 a 4 a 3 Figure 5: A construction o a ive color orest o a given graph using Schnyder woods. Proo. The ollowing construction is illustrated by Fig. 5. Contract the edge a a 3 to a vertex b 3 and the edge a 4 a 5 to a vertex b 4. In these contraction steps the maximal triangles c 5 a a 3 and c a 4 a 5 incident to a, a 3 and to a 4, a 5 are contracted to a single edge, in particular vertices interior to these triangles are removed. Further let b 1 := a 1. This results in an inner triangulation T o the triangle b 1, b 3, b 4. Due to Theorem 4 there exists a Schnyder Wood S o T (we use the colors 1, 3, 4 instead o 1,, 3). Take the colors and orientations o all inner edges not interior to c 5 a a 3 or c a 4 a 5 and not incident to a or a 5 rom T to G. Now color all inner edges incident to a and not interior to c 5 a a 3 in color and orient them towards a, and color all inner edges incident to a 4 and not interior to c a 4 a 5 in color 5 and orient them towards a 5. For the edges interior to c 5 a a 3 construct another Schnyder wood where c 5 has incoming edges in color 5, a 3 has incoming edges in color 3, and a has incoming edges in color. Analogously, construct a Schnyder wood on the edges interior to c a 4 a 5 in the colors, 4, 5. It can easily veriied that this coloring and orientation o the inner edges o G ulills the properties o a ive color orest. In particular, or an inner vertex v not interior to c 5 a a 3 or c a 4 a 5 the properties are still locally ulilled i we replace the outgoing edge o color 3 by an outgoing edge o color or the outgoing edge o color 4 by an outgoing edge o color 5..1 α-orientations Our goal is to connect the setting o ive color orests with the well studied orientations o planar graphs with prescribed outdegrees. Deinition 3. Let H be an undirected graph and α : V (H) N. Then an orientation H o H is called an α-orientation i outdeg(v) = α(v) or all vertices v V (H ). In a ive color orest every inner vertex has outdegree at most 5. The ollowing lemma allows us to add edges so that the outdegree o every inner vertex becomes exactly 5. The statement o the lemma corresponds to the geometric act that in a regular pentagon contact representation o a triangulation G the area between three pentagons corresponding to a ace o G is a quadrilateral with exactly one concave corner. Lemma 1 Let G be endowed with a ive color orest and let be a ace o G that is incident to at most one outer vertex. Then in exactly one o the three interior angles o an outgoing edge is missing in the cyclic order o the respective vertex. 6

7 Figure 6: Full case distinction or Lemma 1 Proo. Since the acial cycle o has length three, it contains an oriented path o length two. For symmetry reasons we can assume that this path is oriented clockwise and the irst edge o this path has color 1. Then because o property (F3) the second edge o this path can only have the colors and 3. Figure 6 shows all possible cases or the orientation and coloring o the third edge o the cycle and veriies the statement or all these cases. Now we deine an extension o G and a unction α 5 such that every ive color orest o G can be extended to an α 5 -orientation o this extension. Deinition 4. The stack extension G o G is the extension o G that contains an extra vertex in every inner ace that is incident to at most one o the outer vertices. These new vertices are connected to all three vertices o the respective ace. We call the new vertices stack vertices and the vertices o G normal vertices. Deinition 5. An orientation o the inner edges o G is a α 5 -orientation i the outdegrees o the vertices correspond to the ollowing values: i v is a stack vertex, α 5 (v) = 5 i v is an inner normal vertex, 0 i v is an outer normal vertex. A ive color orest o G induces an α 5 -orientation o G in a canonical way by keeping the orientation o the edges o G and deining the missing edge o Lemma 1 as the unique incoming edge or every stack vertex. Observation 1 The coloring o the inner edges o a ive color orest can be extended to a coloring o the inner edges o the induced α 5 -orientation that ulills the properties o a ive color orest at all normal vertices. 7

8 . Bijection between ive color orests and α 5 -orientations Now we want to prove that the canonical mapping rom ive color orests to α 5 -orientations is a bijection. For this purpose we need to reconstruct the colors o the inner edges o G i we are given an α 5 -orientation. The idea o this construction will be to start with an inner edge e o G and ollow a properly deined path until it reaches one o the ive outer vertices. Then the color o this outer vertex will be the color o e. This approach is similar to the proo o the bijection o Schnyder Woods and 3-orientations in [4]. Lemma Let C be a simple cycle o length l in G and let all vertices o C be normal vertices, i.e., vertices o G. Then there are exactly l 5 edges pointing rom C into the interior o C. Proo. First we view C as a cycle in G. Let k be the number o vertices strictly inside C. Since G is a triangulation there are exactly k + l aces and 3k + l 3 edges strictly inside C. Now we view C as a cycle in G. In addition to the 3k + l 3 normal edges there are 3 stack edges in each ace, hence, the number o edges in C is 9k + 4l 9. At each stack vertex we see starting edges and at every normal vertex 5. Thereore there are (k + l ) + 5k = 9k + l 4 edges starting at a vertex inside C. Taking the dierence we ind that there are l 5 edges pointing rom a vertex o C into the interior. Next we will show some properties o oriented cycles in ive color orests. By T i we denote the orest consisting o all edges o color i, and by T 1 i we denote the orest T i with all edges reversed. Lemma 3 The orientation T := T i + T i 1 + T i+1 + T 1 i + T 1 i+ o G is acyclic. Proo. Assume there is a simple oriented cycle C o length l in T. Because o the symmetry o the colors in the deinition o a ive color orest, it suices to consider the case that C is oriented clockwise and i = 1. We deine the ollowing auxiliary unction or the ive colors: π(1) := 1, π() := 0, π(3) :=, π(4) := 1, π(5) :=. Let e be an edge o color c ending at vertex v and e an edge o color c starting at v in the orientation T. From Fig. 7 we can read o the ollowing: In the angle o v on the right side between e and the dotted red line there are π(c) edges which are outgoing in the α 5 -orientation o G. In the angle o v on the right side between e and the dotted red line there are π(c ) outgoing edges. Hence, there are exactly π(c ) + ( π(c)) edges pointing away rom v in the right angle between e and e (in the α 5 -orientation o G ). Now let e 1,..., e l be the edges o the cycle C o T and let c i be the color o edge e i. Then the number o edges pointing rom C into the interior (in the α 5 -orientation o G ) is l 1 (π(c i+1 ) π(c i ) + ) + (π(c 1 ) π(c l ) + ) = l. i=1 This is in contradiction to Lemma. Proposition 1 Let C be an oriented cycle in G. Then: (i) C uses at least 3 dierent colors. 8

9 Figure 7: The possible incident edges o a vertex v in T. For an incoming edge o color c below the dotted line the red value is π(c) and counts the number o thick edges in the counterclockwise angle rom this edge to the dotted line. For an outgoing edge o color c above the dotted line the red value is π(c ) and counts the number o thick edges in the clockwise angle rom this edge to the dotted line. The thick edges are exactly those that are outgoing in v in the α-orientation. (ii) I C uses exactly 3 dierent colors, these colors are not consecutive in the cyclic order. (iii) C has two consecutive edges whose colors have distance at most one in the cyclic order. Proo. The irst two statements immediately ollow rom Lemma 3. For the third statement assume that there is an oriented simple cycle C such that the distance o the colors o any two consecutive edges on C is exactly. Because o the symmetry it suices to consider the case that C is oriented clockwise. Let e = uv and e = vw be two consecutive edges on C and let i be the color o e. I the color o e is i +, there is no edge pointing into the interior o C at v. I the color o e is i, there are exactly 4 edges pointing into the interior o C at v. Thereore the total number o edges pointing into the interior o C is even, in contradiction to Lemma. Now we will deine the paths starting with an inner edge e and ending at an outer vertex that allow us to deine the color o e. The idea is to always continue with the opposite outgoing edge, but i we run into a stack vertex, we need to be careul. The paths we will deine are not unique, but we will see that all paths starting with the same edge e end at the same outer vertex. Deinition 6. Let e = uv be an inner edge such that u is a normal vertex. We will recursively deine a set P(e) o walks starting with e by distinguishing several cases concerning v. I v is an outer vertex, i.e., v = a i or some i, the set P(e) contains only one path, the path only consisting o the edge e. I v is an inner normal vertex, let e be the opposite outgoing edge o e at v, i.e., the third outgoing edge in clockwise or counterclockwise direction, and we deine P(e) := {e + P : P P(e )}. 9

10 I v is a stack vertex, let e 1 = vv 1 and e = vv be the let and right outgoing edge o v. Further let e 1 be the second outgoing edge o v 1 ater e 1 in counterclockwise direction and e the second outgoing edge o v ater e in clockwise direction. Note that e i is well deined i v i is not an outer vertex, and that not both o v 1 and v can be outer vertices. I both o e 1 and e are well deined, we deine P(e) := {e+e 1 +P : P P(e 1 )} {e+e +P : P P(e )}. I only e i is well deined, we deine P(e) := {e + e i + P : P P(e i )}. At the moment it is not clear that these walks are inite. I they are inite, they have to end in an outer vertex. But we have to prove that they do not cycle. To be able to apply Lemma to these walks, we will now deine the shortcut o such a walk that avoids the stack vertices. Deinition 7. Let P be a inite subwalk o a walk in P(e) or some edge e that starts and ends with a normal vertex. Then the shortcut P o P is obtained rom P by replacing every consecutive pair uv, vw o edges, where v is a stack vertex, by the edge uw (note that this edge might be oriented wu in the α 5 -orientation). We call uw a shortcut edge. I P is a path with designated start vertex s, then at an inner vertex v we can distinguish edges on the right side o P and edges on the let side o P. We deine r-out P (v) and l-out P (v) to be the number o right respectively let outgoing edges rom path P at vertex v. Lemma 4 Let P be a shortcut walk o length l that does not start and does not end with a shortcut edge. Then the number o edges pointing rom the interior vertices o P to the right (let) o P is (l 1). Proo. Since right and let are symmetric it is enough to prove the lemma or right outgoing edges. The proo is by induction on the length l o the walk. Let P = v 0, e 1, v 1,..., e l, v l. I e l 1 is not a shortcut edge, the statement ollows by induction, because the vertex between two consecutive normal edges o a walk in P(e) has outgoing edges on either side. Now assume that e l 1 is a shortcut edge and let e = v l w, e = wv l 1 be the corresponding edges o the original path P. Let Q be the subwalk o P starting at v 0 and ending at v l extended by the edge v l w. Note that we can apply the induction hypothesis to Q by pretending that w is a normal vertex. Hence, we know that the number o edges pointing rom Q to the right is (l ). The edges v l w and v l v l 1 are consecutive in the cyclic order on incident edges o v l. Deine a sign σ to be 1 i v l v l 1 is right o v l w and +1 otherwise. Let δ = 0 i σ = 1 and v l v l 1 is outgoing at v l 1, otherwise δ = 1. When comparing outgoing edges at v l in P and Q the contribution o δ will account or the edge v l v l 1 i σ = 1 and or the edge v l w i σ = +1. It ollows that r-out P (v l ) = r-out Q (v l ) + σδ, see Fig. 8. Claim 1 r-out P (v l 1 ) = σδ. Proo. I σ = +1, then we are in the v 1 case o the stack vertex part in Deinition 6, i.e., between v l v l 1 and v l 1 v l there is one outgoing edge on the right at v l 1. The claim ollows because in this case δ = 1. I σ = 1, then we are in the v case o the stack vertex part in Deinition 6, i.e., between v l v l 1 and v l 1 v l there is one outgoing edge on the let at v l 1. Now it depends on whether the edge v l v l 1 is outgoing at v l or at v l 1. In the irst case we have δ = 0 and r-out P (v l 1 ) =, in the second case δ = 1 and r-out P (v l 1 ) = 3. In either case this is what the claim says. 10

11 σ = 1 δ = 1 σ = 1 δ = 0 σ = +1 δ = 1 v l v l v l w v l 1 w v l 1 v l 1 w v l v l v l Figure 8: Right outgoing edges at v l and v l 1. The number o right outgoing edges o P is obtained as the sum o those o Q, which is (l ), with r-out P (v l ) r-out Q (v l ), which is σδ, and r-out P (v l 1 ), which is σδ by Claim 1. Hence, the number o right outgoing edges o P is (l 1). Lemma 5 (i) The walks P P(e) are paths, i.e., there are no vertex repetitions in P. (ii) Let P 1, P P(e) be two paths starting with the same edge e. Then P 1 and P end in the same outer vertex. (iii) Let v be a normal vertex and let e 1 = vv 1, e = vv be two dierent outgoing edges at v. Further let P 1 P(e 1 ) and P P(e ) be two paths. Then P 1 and P do not cross and they end in dierent outer vertices. Proo. For (i) assume that P cycles. Let C be a simple cycle which appears as a subwalk o the shortcut o P, and let l be its length. According to Lemma 4, there are at least (l 1) edges pointing into the interior o C. This is in contradiction to Lemma which states that there are only l 5 edges pointing into the interior o C. For (ii) assume that P 1 and P coincide up to a vertex v, then P 1 goes to the let and P to the right. Note that v has to be a stack vertex and let v and v be its predecessors in P 1 and P, i.e., v, v, v appear in this order on both paths. I P 1 and P start in v, we can use a dummy edge v v which makes sure that r-out P1 (v) = l-out P (v) =. For i = 1, let P i be the shortcut o P i. Claim 1 r-out P 1 (v) + l-out P (v) = 6 δ, where δ {0, 1} evaluates to one i the edge vv o P 1 is an outgoing edge at v. Proo. Every outgoing edge o v except possibly the edge vv is a right edge with respect to P 1 or a let edge with respect to P. The edges rom v to v is both, see Fig. 9 (let). Claim I ater splitting at v the two shortcut paths P 1 and P meet at some vertex w which is not an outer vertex, then the irst vertex w ater w is the same on P 1 and P. Proo. For i = 1, let Q i be the subpaths o P 1 and P starting at v and ending at w. At the beginning these paths are extended by an edge v v which may come rom P i, however, we need that the edge is pointing rom v to v. I necessary v may be chosen as a dummy vertex and v v 11

12 w w 1 Q 1 v Q w v Q v Q 1 Figure 9: Overcounts at the two ends o the paths Q 1 and Q. Edges in the green area are counted as right edges o Q 1, those in the red area as let edges o Q. as a dummy edge. At the end the paths are extended by the successor w 1 respectively w o w on P i. I w i is a stack vertex, we can pretend that it is a normal vertex. Let l i be the length o Q i. From Lemma 4 we know that there are exactly (l 1 1) edges pointing rom Q 1 to the right and exactly (l 1) edges pointing rom Q to the let. Let C be the cycle ormed by the two shortcuts Q 1 and Q between v and w. The length o C is l 1 +l 4 and thereore, by Lemma, there are exactly (l 1 + l 4) 5 edges pointing into the interior o C. The sum o the right edges o Q 1 and the let edges o Q correctly accounts or the edges pointing into the interior o C at all vertices except at v and w. The orientation o v v and Claim 1 imply that at v we overcount by exactly 5. It ollows that the overcount at w is ((l 1 1) + (l 1)) ((l 1 + l 4) 5) 5 = 4. This means that each edge except one contributes to the overcount at w, see Fig. 9 (right). Hence, ww 1 and ww have to be identical and w = w 1 = w. Now suppose that P 1 and P end at dierent outer vertices a i and a j. Let v be the last common vertex o the two paths, v has to be a stack vertex. Assume that P 1 goes to the let at v and P goes to the right. Let v be the predecessor o v in P 1 and P. Further let Q 1 and Q be the subpaths o P 1 and P starting at v and ending at a i and a j, extended by a normal edge v v. Again this may be the edge rom P i, however, relevant to us is that it is pointing rom v to v. For i = 1, let Q i be the shortcut o Q i and let l i + 1 be the length o Q i. Let l 3 be the length o the path P 3 between a i and a j that alternates between outer and inner normal vertices. Note that P 3 has l 3 inner vertices and l outer vertices. Let C be the simple cycle in the union o Q 1, Q and P 3. Note that P 3 can have at most one common edge with each Q i. Let ξ {0, 1, } be the number o common edges o P 3 with Q 1 Q. The length o C is l 1 + l + l 3 ξ. I ξ = 0 the edges pointing into C can be obtained by adding the right edges o Q 1 and the let edges o Q subtracting 5 or the overcount at v and adding 3 or each normal vertex o P 3. When Q 1 shares the last edge with P 3, then we have to disregard the contribution o the irst normal vertex w o P 3 and subtract one or the edge wa i+1 which belongs to C and is counted as a right edge o Q 1. Hence, we obtain the ollowing estimate or the number o edges pointing into C: l 1 + l 5 ξ + 3( l 3 ξ) = (l 1 + l + l 3 ξ) 5 l 3. 1

13 This is less than the number o edges pointing into C which is (l 1 +l +l 3 ξ) 5 by Lemma. For (iii) assume that P 1 and P have a common vertex dierent rom v and let w be the irst normal vertex o this kind (note that P 1 and P might already meet at a stack vertex immediately beore w). Let Q 1 and Q be the subpaths o P 1 and P that begin with the predecessors v 1 respectively v o v and end with the successors w 1 respectively w o w. As beore we may replace the true predecessors o v by dummies to have the edges oriented rom v i to v and be sure that r-out Q1 (v) = l-out Q (v) =. We also consider w 1 and w as normal vertices. Let Q 1 and Q be the corresponding shortcut paths, and let l 1 and l be their lengths. Note that the successor edge o v in Q i and Q i is the same. Let C be the cycle we get by gluing the parts o Q 1 and Q between v and w together. Let s be the number o outgoing edges o v between Q 1 and Q interior to C. We assume that looking rom v into the interior o C the let edge o C belongs to Q 1 and the right one to Q. The number o outgoing edges between v v and v 1 v is 5 (r-out Q1 (v) + l-out Q (v) s) = s + 1. The length o C is l 1 + l 4 and thereore, due to Lemma, exactly (l 1 + l 4) 5 edges are pointing into the interior o C. I we add the number o edges pointing rom Q 1 to the right and the number o edges pointing rom Q to the let, we overcount by 5 (s + 1) at v. Hence, the overcount at w must be ((l 1 1) + (l 1)) (5 (s + 1)) ((l 1 + l 4) 5) = 5 + s. To get an overcount 5 at w we need to have every edge in the union o the right edges o Q 1 and let edges o Q. In particular ww 1 is to the let o Q. This implies that at least one o the edges o Q 1 and Q ending in w must be a shortcut edge. It ollows that w is not an outer vertex. Further there are exactly s outgoing edges o w between ww 1 and ww. Thereore we can inductively repeat the argument or the subpaths o P 1 and P starting at w. Now we are able to prove the the main result o this subsection. Theorem 6 The canonical map rom ive color orests to α 5 -orientations is a bijection. Proo. Let F be the set o ive color orests o G and A the set o α 5 -orientations o G. Further let χ : F A be the canonical map and ψ : A F the map that keeps the orientation o the edges as in the α 5 -orientation and colors every edge e in the color o the end vertex o the paths in P(e). Claim 1 The map ψ : A F is well-deined. Proo. Lemma 5 (i) and (ii) show that ψ is a well-deined coloring o the edges o G. It remains to show that this coloring ulills the properties o a ive color orest. Property (F1) is clear rom the construction. Now consider property (F). Because o Lemma 5 (iii) the circular order o the colors o the outgoing edges o an inner vertex has to coincide with the order o the colors o the outer vertices. That the incoming edges o color i are opposite o the outgoing edge o the same color i, ollows rom the construction o the paths. For showing property (F3) assume that there is an inner vertex v with no outgoing edges o colors i and i + or some i. In the α 5 -orientation these missing outgoing edges correspond to edges ending in stack vertices. In the interval between these two outgoing edges there has to be at least one edge e = vw with a normal vertex w. And because o property (F) this edge can only be an incoming edge and the color is i. 13

14 Claim The unction ψ : A F is injective. Proo. We show that we can recover the edges o G rom the ive color orest on G. The orientation at normal vertices can directly be read o rom the ive color orest. Lemma 1 implies that the orientation at stack edges is also prescribed by the ive color orest. Claim 3 The unction χ : F A is injective. Proo. We show that we can recover the coloring o the edges o a ive color orest rom the orientations o the edges, i.e., rom the α 5 -orientation in the image o χ. Clearly, the colors o the edges incident to the outer vertices are known. Moreover, because o property (F) the knowledge o the color o a normal edge incident to an inner normal vertex v implies the knowledge o the colors o all edges incident to v. Since G is connected, this implies that the colors o all edges are unique and known. Since A and F are inite sets, and χ ψ is the identity map on α 5 -orientations, we obtain rom Claims and 3 that ψ and χ and inverse bijections..3 The distributive lattice o ive color orests It has been shown in [6] that the set o all α-orientations o a planar graph carries the structure o a distributive lattice. We need some deinitions to be able to describe the cover relation o this lattice. Deinition 8. A chordal path o a simple cycle C is a directed path consisting o edges interior to C whose irst and last vertex are vertices o C. These two vertices are allowed to coincide. Deinition 9. A simple cycle C is an essential cycle i there is an α-orientation X such that C is a directed cycle in X and has no chordal path in X. Theorem 7 ([6]) The ollowing relation on the set o all α-orientations o a planar graph is the cover relation o a distributive lattice: An α-orientation X covers an α-orientation Y i and only i X can be obtained rom Y by the reorientation o a counterclockwise oriented essential cycle in Y. The reorientation o a counterclockwise (clockwise) oriented essential cycle is called a lip (lop). The ollowing theorem gives a ull characterization o the lip operation in the lattice o ive color orests. Theorem 8 The set o all α 5 -orientations on G carries the structure o a distributive lattice. The lip operation in this lattice is the reorientation o a counterclockwise oriented acial cycle. Proo. Let X be an α 5 -orientation and let C be an essential cycle in G which is directed counterclockwisely, i.e., ready or a lip. Claim 1 There is no edge pointing into the interior o C. Proo. Assume that there is an edge e pointing rom a normal vertex into the interior o C. Let P P(e) be a directed path starting with the edge e and ending in an outer vertex o G. Then P has to cross C at some point and the subpath o P that ends at the irst crossing vertex with C is a chordal path o C, contradicting that C is essential. 14

15 I e = vw is a stack edge, then v is a stack vertex and w a normal vertex. Take any outgoing edge e o w, then a path P P(e ) has to cross C. Together with e this yields a chordal path. Claim The cycle C contains at least one stack vertex. Proo. Assume that C contains only normal vertices. Then according to Lemma there are exactly l(c) 5 0 edges pointing into the interior o C, in contradiction to Claim 1. Now let v be a stack vertex on C. Let w 1 be the predecessor and w be the successor o v on C. We know that the other outgoing edge o v has to point to the outside o C. Now, unless C is a acial cycle the edge w 1 w is an interior chord o C. In either orientation the edge orms a chordal path, hence, C is not essential. Fig. 10 shows the eect o a lip in terms o contacts o pentagons, the eect on the ive color orest can be read rom the igure. Figure 10: A lip o a acial cycle in an α 5 -orientation and its eect on contacts o pentagons. The red segment contributes to another pentagon. 3 The Algorithm In this section we will propose an algorithm to compute a regular pentagon contact representation o a given graph G. First we describe the deinition o a system o linear equations related to a given ive color orest F o G. I the ive color orest is induced by a regular pentagon representation, the solution o the system allows to compute coordinates or the corners o pentagons in this representation. Every inner vertex v o G gets a variable x v representing the side length o the corresponding pentagon and every inner ace gets our variables x (1),..., x (4) representing the segment lengths o the corresponding quadrilateral in clockwise order where the concave corner is located between the edges corresponding to x (1) and x () (see Fig. 11). For the ive inner aces which are incident to two outer vertices o G we add the equation x (1) = 0 since these aces are represented by triangles, not by quadrilaterals. With every inner vertex v we associate ive equations, one or each side. Each o these equations states that the side length x v is equal to the sum o the lengths o the boundary segments o aces incident to the side. More ormally, or i = 1,..., 5, let δ i (v) denote the set o aces o G incident to v in the interval between the outgoing edges o colors i + and i. Then we 15

16 π/5 x (3) x (4) π/5 x () π/5 x (1) Figure 11: The variables or an inner ace. can write these ive equations as x v = δ i (v) x (j v,,i ) with the j v,,i {1,..., 4} appropriately chosen. For geometric reasons the three convex corners o the quadrilaterals corresponding to the aces o G are always exactly (1/5)π. This implies the ollowing two equations or every inner ace x (3) = x (1) + φx (), x (4) = φx (1) + x (). (1) Here φ = 1+ 5 denotes the golden ratio. Finally, we add one more equation to the system which implies that the sum o the lengths o the ace edges building the line segment corresponding to the outer vertex a 1 o G is exactly 1, i.e., δ 1 (a 1 ) x (j a 1,,1) = 1 with j a1,,1 {1,..., 4} appropriately chosen. In the equations δ 1 (a 1 ) x (j a 1,,1) = 1, and δ i (v) x (j v,,i ) x v = 0 we eliminate the variables x (3), x (4) using substitutions according to (1). The resulting system o linear equations is denoted A F x = e 1, here A F is the coeicient matrix depending on the ive color orest F and e 1 is the irst standard unit vector. We next show that the system o linear equations is uniquely solvable. For this purpose we need a lemma about perect matchings in plane bipartite graphs. The lemma is well known rom the context o Paian orientations, see e.g., [17], we include a proo or completeness. Let H be a bipartite graph with vertex classes {v 1,..., v k } and {w 1,..., w k }. Then a perect matching o H induces a permutation σ S k by σ(i) = j : {v i, w j } M. We deine the sign sgn(m) o a perect matching M as the sign o the corresponding permutation. Lemma 6 Let H be a bipartite graph and let M, M be two perect matchings o H. I the symmetric dierence o M and M is the disjoint union o simple cycles C 1,..., C m such that, or i = 1,..., m, the length l i o C i ulills l i mod 4, then sgn(m) = sgn(m ). I H is a plane graph such that each inner ace o H is bounded by a simple cycle o length l mod 4, this property is ulilled or any two perect matchings o H. Proo. For i = 1,..., m, there is an n i N with l i = 4n i +. Then on the vertices o C i the permutation σ corresponding to M and the permutation σ corresponding to M dier in a cyclic 16

17 permutation τ i o length n i + 1. Hence, we have σ = σ τ 1 τ m and thereore sgn(σ ) = sgn(σ) sgn(τ 1 ) sgn(τ m ) = sgn(σ) ( 1) n1 ( 1) nm = sgn(σ). In the case that H is a plane graph such that each inner ace o H is bounded by a simple cycle o length l mod 4, or each cycle o length l with k vertices in its interior the ormula l + k mod 4 is valid. This can be shown by induction on the number o aces enclosed by the cycle. Since each o the cycles C 1,..., C m contains an even number o vertices in its interior, this implies l i mod 4 or i = 1,..., m. Theorem 9 The system A F x = e 1 is uniquely solvable. Proo. We show that det(a F ) 0. Let  F be the matrix obtained rom A F by multiplying all columns corresponding to vertices o G with 1. Since in A F all entries in these columns are nonpositive and the entries in all other columns are non-negative, all entries o  F are non-negative. Further we have det(a F ) = ( 1) n 5 det(â F ). Now we want to interpret the Leibniz ormula o det(â F ) as the sum over the perect matchings o a plane auxiliary graph H F. Let H F be the bipartite graph whose irst vertex class v 1,..., v k consists o the variables o the equation system and whose second vertex class w 1,..., w k consists o the equations o the equation system. There is an edge v i w j in H F i and only i ( F ) ij > 0. Then we have det(â F ) = sgn(σ) ( F ) iσ(i) = sgn(m)p M, σ i M where the second sum goes over all perect matchings o H F and where P M > 0 or all perect matchings M. Next we will deine an embedding o H F into the plane. We start with a crossing-ree straightline drawing o G. Then we put pairwise disjoint disks around the inner vertices and cut the bordering circles into ive arcs. These ive arcs are the drawings o the ive equation-vertices incident to the respective vertex and each o these arcs is contained in exactly those aces o the embedding o G which are involved in the corresponding equation. Then every inner ace o G is intersected by exactly our o these arcs, two rom one o the incident vertices and one rom the other two vertices. We denote them by A 1,..., A 4 in clockwise order where A 1 and A come rom the same vertex. We place the vertices corresponding to x (1) and x () inside, but outside o the disks o the three incident vertices o. We connect x (1) to A 1 and A 4, and x () to A and A 3. Up to this point the drawing is crossing-ree. Finally we add the two intersecting edges x (1) A 3 and x () A 4 inside. See Fig. 1 or an illustration. Claim 1 The graph H F has a perect matching. Proo. We describe an explicit construction o a perect matching o H F. The ive equationvertices adjacent to a vertex v o G are corresponding to the ive colors o the ive color orest. We always match the vertex v with the equation-vertex o color 4. The equation-vertices o colors and 3 are matched with one o the two variable-vertices o the last incident ace in clockwise order, and the equation-vertices o colors 5 and 1 are matched with one o the two variable-vertices o the last incident ace in counterclockwise order (see Fig. 13 (let)). Now we will show that each ace (except the ive corner-aces) is matched exactly twice. We call a segment o a acial quadrilateral B a short segment i it is incident to the concave corner, 17

18 Figure 1: Embedding o H F into the plane. 1 Figure 13: Let: Cases or the construction o the perect matching. Right: The complete perect matching in a small instance. and we call it a long segment otherwise. For two adjacent segments o B we call the segment, whose containing pentagon side ends in the contact point o the two segments, the cut segment. Above we mentioned that the ive equation-vertices o a vertex o G correspond to the ive colors. Since each segment o B is involved in exactly one o these equations, we can also associate a color with each o the segments o B. We distinguish three cases concerning segments o color 4 (Fig. 13 (let) shows two o the cases). I B has a short segment o color 4, the other short segment and the cut long segment are matched. I B has a long segment o color 4, the short segment, which is neighboring the segment o color 4, and the cut segment o the two remaining segments are matched. I B has no segment o color 4, or each pair o neighboring long and short segments the cut segment is matched. Hence, in every case the ace is matched exactly twice. Since each ace is matched exactly twice, its two variable-vertices are matched exactly once. It can easily be seen that each corner-ace is matched exactly once, except the corner-ace o color 4 which is not matched. Finally we match the corner-ace o color 4 with the equationvertex corresponding to the non-homogeneous equation, and obtain a perect matching. Figure 13 (right) shows an example. 18

19 Let M 0 be the set o perect matchings o H F that do not contain both edges o any pair o crossing edges. Claim Let M 1, M M 0. Then sgn(m 1 ) = sgn(m ). Proo. The symmetric dierence o M 1 and M is a disjoint union o simple cycles. Let C be one o these cycles. Due to Lemma 6 it suices to show that l(c) mod 4. Now consider a pair v a w b, v c w d o crossing edges that are both contained in C, i.e., v a and v c are the two variable vertices o a ace and w b, w d correspond to the equations or sidelength o dierent pentagons, in particular v a w b v c w d is a 4-cycle in H F. There are two paths P 1, P such that C = v a w b P 1 w d v c P. I C looked like C = v a w b P 1 v c w d P, the cycle v c w b P 1 would have odd length, and that is not possible since H F is bipartite. Let C be the cycle deined by C := v c w b P 1 w d v a P where P denotes the reversed path P. Note that l(c ) = l(c) and that C is contained in the symmetric dierence o the perect matchings M 1 and M obtained rom M 1 and M in the ollowing way: In one o the matchings we replace v a w b by v a w d and in the other matching v c w d by v c w b. Further all edges o P 1 switch the matching. The cycle C has one crossing less than C. By iterating this process we obtain a cycle C with l(c ) = l(c) which is crossing-ree and contained in the symmetric dierence o two perect matchings o H F. Since C is crossing-ree, it is also contained in a spanning subgraph H F o H F that keeps all none crossing edges o H F and exactly one o the two edges o every crossing pair. Then H F inherits a crossing-ree drawing rom H F where every inner ace is a simple cycle o length 6 (note that every ace corresponds to a corner o a pentagon in the pentagon contact representation). Due to Lemma 6 this implies l(c ) mod 4. Now we will consider all perect matchings o H F. We divide them into equivalence classes according to the ollowing equivalence relation: Two perect matchings M, M are equivalent i M can be obtained rom M by exchanging pairs o crossing edges with the two non-crossing edges on the same our vertices (we call them twin edges). Exchanges in both directions are allowed. Note that each equivalence class contains exactly one o the matchings o M 0 we considered in Claim. Let M 0 M 0 be one o these matchings and let k = k(m 0 ) be the number o pairs o twin edges contained in M 0. Because each pair o twin edges can be exchanged independently with the corresponding pair o crossing edges, the equivalence class A M0 o M 0 consists o k matchings and or i = 0,..., k, the class contains ( ) k i matchings with exactly i pairs o crossing edges. In the product P M0 o entries o A F corresponding to the edges o M 0 we see a contribution o φ or each pair o twin edges, see equations (1). The contribution or a pair o crossing edges is 1. Thereore M A M0 sgn(m)p M = k i=0 ( ) ( ) k 1φ i sgn(m 0 )( 1) i P M0 i = sgn(m 0 )P M0 (1 1φ ) k and det(â F ) = M 0 M 0 sgn(m)p M = sgn(m 0 ) P M0 (1 1φ ) k(m0 ) }{{} M A M0 M 0 M 0 >0 }{{} >0 Because o Claim we have sgn(m 0 ) = sgn(m 0 ) or any two matchings M 0, M 0 M 0. Finally this implies det(â F )

20 The ollowing lemma will help us to prove that a non-negative solution o the system A F x = e 1 leads to a regular pentagon contact representation o G. Lemma 7 Let H be an inner triangulation o a polygon. For every inner ace o H with vertices v 1, v, v 3 in clockwise order let T be a triangle in the plane with vertices p(, v 1 ), p(, v ), p(, v 3 ) in clockwise order such that the ollowing conditions are satisied: (i) For each inner vertex v o H with incident aces 1,..., k k β( i, v) = π i=1 where β(, v) denotes the interior angle o T at p(, v). (ii) For each inner edge vw o H with incident aces 1, p( 1, v) p( 1, w) = p(, v) p(, w), i.e., the vector between v and w is the same in T 1 and T. Then there exists a crossing-ree straight line drawing o H such that the drawing o every inner ace can be obtained rom T by translation. Proo. Let H be the dual graph o H without the vertex corresponding to the outer ace o H. Further let S be a spanning tree o H. Then we can glue the triangles T o all inner aces o H together along the edges o S. We need to show that the resulting shape has no holes or overlappings. For the edges o S we already know that the triangles o the two incident aces are touching in the right way. For the edges o the complement S o S we still need to show this. We consider S as a subset o the edges o H. Note that S is a orest in H. Let e be an edge o S incident to a lea v o this orest that is an inner vertex o H. Then or all incident edges e e o v we already know that the triangles o the two incident aces o e are touching in the right way. But then also the two triangles o the two incident aces o e are touching in the right way because v ulills property (i). Since the set o edges we still need to check is still a orest, we can iterate this process until all inner edges o H are checked. Theorem 10 The unique solution o the system A F x = e 1 is non-negative i and only i the ive color orest F is induced by a regular pentagon contact representation o G. Proo. Assume there is a regular pentagon representation S o G that induces the ive color orest F. Then the edge lengths given by S deine a non-negative solution o A F x = e 1. For the opposite direction, assume the solution o A F x = e 1 is non-negative. To be able to apply Lemma 7 we irst construct an internally triangulated extension o the skeleton graph o a hypothetical regular pentagon contact representation with induced ive color orest F. We start with a crossing-ree straight-line drawing o G. Add a subdivision vertex on each edge o G. Moreover, or each inner vertex v draw an edge ending at a new vertex reaching into each ace with a missing outgoing edge o v. Then connect all the new adjacent vertices o v in the cyclic order given by the drawing. At this point inner aces o G are subdivided into our triangles and a quadrangle (shown gray in Fig. 14). One o the vertices o the quadrangle is the new interior vertex w o the ace. Connect vertex w to the subdivision vertex diagonally in the quadrangle (shown dashed in Fig. 14). 0