What Makes a Tree a Straight Skeleton?∗

CCCG 2012, Charlottetown, P.E.I., August 8–10, 2012

What makes a Tree a Straight Skeleton?∗

Oswin Aichholzer† Howard Cheng‡ Satyan L. Devadoss§ Thomas Hackl† Stefan Huber¶

Brian Li§ Andrej Risteskik

Abstract

Let G be a cycle-free connected straight-line graph with predefined edge lengths and fixed order of incident edges around each vertex. We address the problem of deciding whether there exists a simple polygon P such that G is the straight skeleton of P . We show that for given e G such a polygon P might not exist, and if it exists it f(e) might not be unique. For the later case we give an example with exponentially many suitable polygons. For small star graphs and caterpillars we show necessary and sufficient conditions for constructing P . Figure 1: The straight skeleton (thin) of a simple poly- Considering only the topology of the tree, that is, gon (bold) is defined by the propagating wavefront (dot- ignoring the length of the edges, we show that any tree ted). whose inner vertices have degree at least 3 is isomorphic to the straight skeleton of a suitable convex polygon. wavefront edge and splits the wavefront into pieces, see 1 Introduction Figure 1. The straight skeleton (P ) is defined as the set of loci that are traced out byS the wavefront vertices The straight skeleton (P ) of a simple polygon P is a and it partitions P into polygonal faces. Each face f(e) skeleton structure like theS Voronoi diagram, but consists belongs to a unique edge e of P . Each straight-skeleton of straight-line segments only. Its definition is based edge belongs to two faces, say f(e1) and f(e2), and lies on a so-called wavefront propagation process that corre- on the bisector of e1 and e2. sponds to mitered offset curves. Each edge e of P emits Straight skeletons have many applications, like auto- a wavefront that moves with unit speed to the interior mated roof construction, computation of mitered offset of P . Initially, the wavefront of P consists of parallel curves, topology-preserving collapsing of areas in geo- copies of all edges of P . However, during the wavefront graphic maps, or solving fold-and-cut problems. See [4] propagation, topological changes occur: An edge event and Chapter 5.2 in [3] for further information and de- happens if a wavefront edge shrinks to zero length. A tailed definitions. split event happens if a reflex wavefront vertex meets a Although straight skeletons were introduced to computational geometry in 1995 by Aichholzer et al. [1], ∗A preliminary version of this paper appeared at Eu- their roots actually go back to the 19th century. In roCG 2012. Research of O. Aichholzer partially supported textbooks about the construction of roofs (see e.g. [6], by the ESF EUROCORES programme EuroGIGA – Com- PoSe, Austrian Science Fund (FWF): I 648-N18. T. Hackl pages 86–122) using the angle bisectors (of the polygon was funded by the Austrian Science Fund (FWF): P23629- defined by the ground walls) was suggested to design N18. S. Huber was funded by the Austrian Science Fund roofs where rainwater can run off in a controlled way. (FWF): L367-N15. H. Cheng, S.L. Devadoss, B. Li, and A. This construction is called Dachausmittlung and became Risteski were funded by NSF grant DMS-0850577. rather popular. See [5] for related and partially more †Institute for Software Technology, Graz University of Tech- involved methods to obtain roofs from the ground plan nology, [oaich|thackl]@ist.tugraz.at of a house. In this book detailed explanations of the ‡University of Arizona, Tucson, AZ 85721, [email protected] constructions and drawings of the resulting roofs can §Williams College, Williamstown, MA 01267, be found. [satyan.devadoss|brian.t.li]@williams.edu Maybe not surprisingly, none of this early work men- ¶Department of Mathematics, UniversitätSalzburg, Austria, [email protected] tions the ambiguity of the non-algorithmic definition of kPrinceton University, Princeton, NJ 08544, the construction. It can be shown that using solely bi- [email protected] sector graphs does not necessarily lead to a unique roof 24th Canadian Conference on Computational Geometry, 2012

construction, and actually does not even guarantee a denotes the degree of the vertex of T that corresponds plane partition of the interior of the defining boundary. to u. Applying this technique recursively — e.g., in a See [1] for a detailed explanation and examples. breadth-first search fashion starting from v — gives us An interesting inverse problem was motivated to us finally a polygon Pm, where m denotes the number of by Lior Pachter and investigations started in [2]: Which inner nodes of T , whose straight skeleton (Pm) is topo- graphs are the straight skeleton of some polygon? In logically equivalent to T by construction.S Furthermore, other words, how can straight skeletons be characterized in our induction step we guarantee that all polygons among all graphs? P1,...,Pm remain convex. For the induction step, we consider a leaf vertex u in 2 Finding straight skeletons of given topology (Pi) and we denote by e the incident straight-skeleton S edge of u. As u is a leaf vertex of (Pi) it is also a poly- First of all, the straight skeleton (P ) of a simple polygon vertex of P . Hence, e lies onS the straight-skeleton S i gon is known to be connected and cycle-free. Hence, faces f(s1) and f(s2) of the two incident polygon edges we can rephrase our question as follows: which trees T s1 and s2 of u. Since Pi is convex by induction, all faces are realized as straight skeletons of simple polygons? At of (P ) are convex, too. Hence, the projection lines of S i first we concentrate on the topological structure of trees the mid point u0 of e onto s1 and s2 are completely con- only and ignore their geometry. tained in f(s1) and f(s2), respectively. Let us consider the circular arc C that is centered at u0 and tangen- Theorem 1 For any tree T , whose inner vertices have tial to s1 and s2 such that C forms a round convex cap at least degree 3, there exists a feasible (convex) polygon of the corner u of Pi, see Figure 2. In the induction P such that (P ) possesses the same topology as T . S step, we locally bevel the corner u of Pi by any convex polygonal chain with k 2 vertices that is tangential Note that within convex polygons the straight skele- ≥ ton and the Voronoi diagram are identical. Hence, by to C. By that the edge e is truncated to u0 and we ob- the above theorem, any tree is also isomorphic to the tain k additional straight-skeleton edges e1, . . . , ek that Voronoi diagram of a suitable convex polygon. are incident to u0, as desired. The resulting polygon Pi+1 is again convex and a locally beveled version of Pi. Proof. We first choose any inner vertex v of T to be Note that the remaining straight skeleton of Pi remains the root of T . Then we construct a regular polygon P1 unchanged for Pi+1. with d(v) sides, where d(v) denotes the degree of v. The straight skeleton (P1) is a star graph comprising one inner node and dS(v) incident edges, which correspond 3 Abstract geometric trees to v and its incident edges in T . It remains to attach the corresponding subtrees from T to each leaf vertex The original problem motivated by Lior Pachter and for which investigations started in [2] does not only ask for of (P1), if there are any. InS the remainder of the proof we describe an induc- a specific topology of (G), but also asks for certain geometric requirementsS that are to be fulfilled by (P ). tive step by which we locally transform a polygon Pi to In particular, we want to find a polygon P for whichS (i) Pi+1 in order to attach to a leaf vertex u of (Pi) a miss- ing number k = d(u) 1 of incident edges,S where d(u) (P ) has a specific topology, (ii) the edges of (P ) have − aS specific length and (iii) the cyclic order ofS incident edges at vertices of (P ) is given. u S To give a more formal problem definition we denote with abstract geometric graphs the set of combinatorial e graphs, where the length of each edge and the cyclic or- Pi der of incident edges around every vertex is predefined

+1 (and cannot be altered). Let be the set of cycle- Pi G C free connected abstract geometric graphs. Denote with e E(G) an embedding of G in the plane, that is, 1 e2 ∈ G ek the vertices of G are points in R2 and the edges of G are straight-line segments of the predeﬁned length, s s1 2 connecting the corresponding points and respecting the predeﬁned cyclic order of incident edges around each u0 e vertex. Further, denote with PE(G) the polygon resulting from connecting the leaves of G (with straight-line Figure 2: The induction step in order to add k edges to segments) in cyclic order for the embedding E(G). We a terminal vertex of (Pi), with k = 3, by beveling the call a simple polygon P suitable if its straight skele- S E(G) corner u of Pi. ton (P ) = E(G), for the embedding E(G). If there S E(G) CCCG 2012, Charlottetown, P.E.I., August 8–10, 2012

v = v v gi 1 2 i 1 i − −

li αi u v1 v t ≡ n+1 Figure 3: Example of a feasible cycle-free connected abstract geometric graph G (leaves of G are shown as white dots). Left: Arbitrary embedding E(G) and (non- vn simple) polygon PE(G) (dotted). Right: Suitable poly- Figure 4: Construction of L (t, A) (and E(S )) for a gon PE0(G) for a different embedding E0(G), which is Sn n given S and a fixed distance t and assignment A. equal to (PE0(G)). A set of wavefronts of PE0(G) at n different pointsS in time are depicted in gray.

be convex or reflex, as seen from u. As l1 = maxi li, exists a suitable polygon for a graph G , we call G v1 and vn+1 are always convex (fact (3) in Observa- feasible, see Figure 3. ∈ G tion 1). For the remaining vertices any convex/reflex The obvious questions which arise from these defini- assignment, which respects the facts (2) and (3) in Ob- tions are: Which graphs of are feasible? Are the suit- servation 1, can be considered. The edges of LSn (t, A) able polygons for feasible graphsG unique modulo rigid have equal orthogonal distance t to u. Of course, not all motions? How can one construct a suitable polygon for possible combinations of t and an arbitrary embedding a feasible graph? E(Sn) allow such a polyline, but it is possible to con-

struct LSn (t, A) and E(Sn) simultaneously for a fixed t (0, min l ]. 3.1 Star graphs ∈ i i For a fixed assignment A and a fixed t (0, min l ] we ∈ i i All polygon edges whose straight-skeleton faces con- construct LSn (t, A) (and E(Sn)) in the following way. tain a common vertex u (of the straight skeleton) have Consider the circle C with center u and radius t. Start equal orthogonal distance t to u, because their wave- with v1 at polar coordinate (l1, 0), with u as origin. For front edges reach u at the same time t. That is, the each vi, i = 2 ... (n + 1), consider a tangent gi 1 to C − supporting lines of those polygon edges are tangential (such that the vertices will be placed counter-clockwise to the circle with center u and radius t. Thus, in this around the circle) through vi 1. If vi 1 is convex, then − − section we consider a subset of , the so called star there exist two points with distance li (l1 for vn+1) on G graphs. A star graph S , for n 3 has (n + 1) ver- gi 1. If vi is assigned to be reflex, then vi is placed n ∈ G ≥ − tices, one vertex u with degree n and n leaves v , . . . , v on the point closer to vi 1, and if vi is assigned to be 1 n − ordered counter-clockwise around u. The length of each convex, then vi is placed on the other point. If vi 1 − edge uv , with 1 i n, is denoted by l . W.l.o.g. let is reflex, then there exists only one applicable point for i ≤ ≤ i l = max l . Observe that the polygon P is star placing vi on gi 1. See Figure 4. 1 i i E(Sn) − shaped and vivi+1 (with vn+k := v1+(k 1) mod n) are its The LSn (t, A) constructed this way is unique (for edges. − fixed t and A), and may be not simple (e.g. when cir- cling C many times), simple but not closed (v v ), n+1 6≡ 1 Observation 1 If Sn is a feasible star graph and or simple and closed (vn+1 v1). In the latter case, the ∈ G construction reveals a witness≡ pair (t, A) for the exis- PE(Sn) is a suitable polygon of Sn, then (1) all straight- skeleton faces are triangles, (2) two consecutive vertices tence of some E(Sn), a suitable polygon PE(Sn), and vi, vi+1 can not both be reflex, (3) li < li 1 for each re- thus the feasibility of Sn. ± flex vertex vi of PE(Sn), and (4) all edges of PE(Sn) have It is easy to see that for each suitable polygon PE(Sn), equal orthogonal distance t to u, with t (0, min l ]. there exists a polyline LS (t, A) (just duplicate the ver- ∈ i i n tex v1). Hence, deciding feasibility of Sn is equivalent to As a given S is possibly not feasible and a suit- finding an assignment A and a t (0, min l ] such that n ∈ G ∈ i i able polygon may not be known or might not exist, we LSn (t, A) is closed and simple. For a polyline LSn (t, A) define a polyline LSn (t, A): The vertices v1, . . . , vn+1 of and a corresponding embedding E(Sn), we denote with

LSn (t, A) are the leaves, v1, . . . , vn, of Sn, in the same αi, i = 1 . . . n, the counter-clockwise angle at u, spanned order as for Sn, and one additional vertex vn+1 suc- by uvi and uvi+1. Note that for a suitable polygon ceeding v . The vertices v , . . . , v , v have the cor- P α can be deﬁned the same way, with v v . n 1 n n+1 E(Sn) i n+1 ≡ 1 responding distances (predeﬁned in Sn) l1, . . . , ln, l1 to It is easy to see that the sum of all αi is 2π if and only u. A is an assignment for each vertex whether it should if LSn (t, A) is closed and simple. 24th Canadian Conference on Computational Geometry, 2012

Lemma 2 Let Sn , distance t (0, mini li] and as- convex or all but v5 convex. It is easy to check that for ∈ G ∈ P signment A be fixed, and let LSn (t, A) be the resulting both assignments i αi > 2π, for every t (0, mini li]. Pn ∈ polyline. Then αA(t) := i=1 αi can be expressed as To prove the second claim consider a star graph with n = 5, l1 = l3 = 1, l2 = 0.6, l4 = 0.79, and l5 = 0.75. n n X t X t Assign all vertices convex, except for v . Then P α α (t) = 2 arccos 2 arccos . (1) 2 i i A l − l evaluates to 2π for t 0.537 and t 0.598. Hence, i=1 i i=1 i ≈ ≈ vi convex vi reflex there exist (at least) two different suitable polygons for this star graph. Proof. Recall that v1 and vn+1 are convex by the assumption l1 = ln+1 = maxi li. It is easy to see that Note that for the latter example two suitable polygons α = arccos t +arccos t if v is convex and v is con- exist that even share the same reflexivity assignment. In i li li+1 i i+1 vex, α = arccos t arccos t if v is convex and v the following we discuss sufficient and necessary condi- i li li+1 i i+1 − t t tions for the feasibility of a star graph S4. By Lemma 3 is reflex, and αi = arccos + arccos if vi is reflex − li li+1 we know in which cases suitable convex polygons exist. and vi+1 is convex. If vi is reflex then αi 1 +αi sums up − The remaining cases are solved by the following lemma. to arccos t + arccos t + arccos t + arccos t li−1 li li li+1 − t Lemma 6 Consider an S4 for which no suitable convex 4 arccos , because vi 1 are both convex (fact (2) in Ob- li ± polygon exists. A suitable non-convex polygon exists if servation 1). Thus, summing over all α results in the i and only if 1 < P 1 . min l j=1,lj =mini li l claimed formula. i i 6 j Proof. First of all, if a polyline has two or more re- We define a suitable polygon to be unique if it is the flex vertices assigned then α (t) < 2π, as each positive only suitable polygon modulo rigid motions. For the A summand in Equation (1) is bound by π/2. Hence, we following result we use the first derivative of α : A only need to consider polylines with exactly one reflex n n X 1 X 1 vertex, which implies αA(0) = 2π. α0 (t) = 2 2 . (2) For simplicity, we may reorder v and l such that l = A p 2 2 p 2 2 i i 4 li t − li t i=1 i=1 mini li. We show that for suitable non-convex polygons vi reflex − vi convex − v4 needs to be reflex. Assume to the contrary that some Lemma 3 A suitable convex polygon for a star graph vk, with 1 k 3, is reflex. In this case we obtain P mini li ≤ ≤ 1 2 2 1 2 2 Sn exists if and only if arccos π. If a suit- that α0 (t) < 0 as /√l4 t dominates /√l t for all i li A − k− able convex polygon exists then it is unique.≤ t [0, l ). But since α (0) = 2π we see that α (t) < 2π ∈ 4 A A for all t (0, mini li]. Proof. As all vertices are assumed to be convex, we Observe∈ that the assumption in the lemma, that no obtain αA(0) = nπ > 2π. Furthermore, we observe suitable convex polygon exists, is equivalent to αA(l4) > that αA(t) is monotonically decreasing since α0 (t) < 0 A 2π. Recall that αA(0) = 2π. Hence, if αA0 (0) < 0 then for all t (0, mini li]. Hence, there is a t (0, mini li] there exists a t (0, l ) such that α (t) = 2π, as α is ∈ ∈ ∈ 4 A A with α(t) = 2π if and only if αA(mini li) 2π which is continuously differentiable. P mini li ≤ i arccos π. If this is the case the solution is Finally, we show that if α (0) 0 then α (t) > 0 li ≤ A0 A0 unique as α(t) is monotonic. for all t (0, l ). Hence, there is no≥t (0, l ] such that ∈ 4 ∈ 4 αA(t) = 2π. From Equation (2) we get that αA0 (t) > 0 For n = 3, αA(0) = 3π and αA(mini li) < 2π, and is equivalent to thus we immediately get the following corollary. s 3 3 2 2 1 X 1 X li l4 Corollary 4 For every S3 there exists a unique suitable > 1 > 1 − p 2 2 p 2 2 2 2 l t l t ⇔ − li t convex polygon. 4 − i=1 i − i=1 − Considering star graphs with n = 5, we show in the The right side of this equivalence is true since following lemma that they are not always feasible, and 3 s 3 s X l2 l2 X l2 l2 that suitable polygons (if they exist) are not always 1 1 i − 4 > 1 i − 4 , (3) ≥ − l2 − l2 t2 unique. i=1 i i=1 i −

where the ﬁrst inequality is given by α0 (0) 0 and the Lemma 5 There exist infeasible star graphs, Sn . A ∈ G second inequality holds for all t (0, l ). ≥ Further, there exist feasible star graphs for which mul- ∈ 4 tiple suitable polygons exist. To conclude, we have shown that if no suitable convex polygon exists for some S4, then a suitable non-convex Proof. To prove the ﬁrst claim consider a star graph polygon exists for this S4 if and only if α0(0) < 0, which is equivalent to 1 < P 1 , as claimed with n = 5, l1 = l2 = l3 = l4 = 1, and l5 = 0.25. There min l j=1,lj =mini li l i i 6 j exist only two possible assignments: either all vertices in the lemma. CCCG 2012, Charlottetown, P.E.I., August 8–10, 2012

3.2 Caterpillar graphs

The techniques developed in the previous section can be i v1 generalized to so-called caterpillar graphs. A caterpillar π π βi + βi graph G is a graph that becomes a path if all its li 2 − 2 2 li leaves (and∈ G their incident edges) are removed. We call 1 this path the backbone of G. Figure 3 shows a caterpillar i i i i+1 v l3 l v graph whose backbone comprises three backbone edges. 3 ki 0 i In general, a caterpillar graph has m backbone ver- v0 tices, consecutively denoted by v1, . . . , vm. We denote P 0 0 ri i0 i r the adjacent vertices of a backbone vertex v0, with ki li i+1 i+1 ki incident edges, by vi , . . . , vi , such that vi = v for i 1 ki ki 0 vj βi i 1 i < m. Furthermore, we denote by lj the length of arcsin ri e ≤ li i i j r r the edge v0vj, see Figure 5. Let us consider a polygon P i+1 − i whose straight skeleton (P ) forms a caterpillar graph. S Figure 5: A section of a polygon P for which (P ) is a Observation 2 All edges of P whose straight-skeleton caterpillar graph. S i faces contain the same backbone vertex v0 have identical i orthogonal distance to v0. Corollary 8 The sum of the inner angles of P with We denote this orthogonal distance by r . Hence, the E(G) i convexity assignment A is a function supporting lines of the corresponding polygon edges are i ( rv tangents to the circle of radius ri centered at v0, see n arcsin j v is convex X lj j Figure 5. αA(r1) = 2 rvj , (4) π arcsin vj is reflex j=1 − lj Lemma 7 The radii r2, . . . , rm of a suitable polygon P for some given caterpillar graph G are determined E(G) where rvj denotes the radius of the circle at the backbone by r1 and the predefined edge lengths of G according to vertex that is adjacent to vj and lj denotes the length of the following recursions, for 1 i < m: the incident edge of G. ≤ r = r + li sin β i+1 i ki i The previous corollary provides us with a tool in or- k βi = βi 1 + (1 i/2)π+ der to find suitable polygons P for caterpillar graphs − − E(G)  r G. We know that for any suitable polygon P the ki 1 i i is convex E(G) X− arcsin li vj j identity αA(r1)=(n 2)π must hold. Hence, we can de- ri i is reflex − π arcsin li vj termine all suitable polygons PE(G) as follows: for all j=1 − j vi =vi−1 n j 6 0 2 possible assignments A we determine all r1 such that 1 0 αA(r1)=(n 2)π. For i = 1 we define that β0 = 0 and vj = v0 being true − 6 For any such pair (A, r1) we construct a polyline for all 1 j < k1. ≤ v1, . . . , vn, vn+1 by a similar method as outlined for star Proof. Denote with e one of the two edges of PE(G) graphs: shooting rays tangential to circles centered at i i+1 the backbone vertices vi . In order to switch over from whose faces of (PE(G)) contain the edge v0v0 . The 0 S i vi to vi+1, we consider the previously constructed ray, supporting line of e is tangential to the circles at v0 and 0 0 i+1 which needs to be tangential to the two circles centered v0 . Considering the shaded right-angled triangle in i at both, vi and vi+1, respectively. As the length of the Figure 5, we obtain ri+1 ri = lk sin βi. 0 0 − i · edge vi vi+1 is given, the center vi+1 of the next cir- Consider the polygon Pi0 (bold in Figure 5) which 0 0 0 comprises the edges of PE(G) whose faces of (PE(G)) cle is uniquely determined, cf. Figure 5. If there is any i S non-backbone edge with length li < r then there is no contain v0, trimmed by two additional edges orthogonal j i i 1 i i i+1 suitable polygon for that particular pair (A, r ). For to v0− v0 and v0v0 , respectively. Pi0 comprises ki+2 1 each candidate polyline we check whether it is closed, vertices (k1+1 for P10) and hence, the sum of inner angles simple and forms a suitable polygon. Note that all suit- equals kiπ ((k1 1)π for P10). On the other hand, we can − able polygons can be constructed by the above method. express this sum as follows (also for P10), which implies the second recursion: Lemma 9 There is at most a finite number of suitable kiπ = 2π + 2βi 1 2βi+ polygons P for a caterpillar graph G. − − E(G)  r ki 1 arcsin i vi is convex X−  li j Proof. As α is analytic, there are no accumulation 2 j A ri i points in the set r : α (r ) = (n 2)π . Otherwise, π arcsin li vj is reflex 1 A 1 j=1 − j { − } vi =vi−1 αA would be identical to (n 2)π. In other words, j 6 0 − 24th Canadian Conference on Computational Geometry, 2012

there is only a finite number of possible pairs (A, r1) 4 Conclusion that correspond to a suitable polygon. In this work, we considered the inverse problem of com- Lemma 10 There exists a caterpillar graph with 3m puting the straight skeletons of simple polygons. First, m 2 we proved that each tree, whose inner vertices have de- vertices having 2 − suitable polygons. gree at least 3, can be realized as the straight skeleton of a convex polygon. The constructive proof also pro- Proof. We consider a caterpillar graph with m back- vides the outline for an algorithm to construct a suitable bone vertices, for which the two outer backbone ver- polygon. tices are of degree three and the m 2 inner backbone Next, we considered a more restrictive version of vertices are of degree four. We set the− length of the non- the original question by predefining the lengths of backbone edges to √2/4 and the length of the backbone k k+1 k 1 the straight-skeleton edges and their circular orders at edge e = v v to 3/4 2 . We embed the backbone k 0 0 − straight-skeleton vertices. For star graphs we showed as a rectilinear path and· at each vk, with 1 < k < m, we 0 that there exists a unique suitable convex polygon for either make a left or right turn from e to e . This k 1 k every S . Further, we derived that for general star gives us 2m 2 possible embeddings, see− Figure 6. We 3 − graphs feasibility is neither guaranteed nor unique, and now consider the polygon P shown in Figure 6, which we gave sufficient and necessary conditions for the ex- forms a rectilinear hose (a mitered offset curve) around istence of a suitable polygon for all S . Furthermore, the embedding of G with thickness 1 . It remains to 4 2 we gave a simple necessary and sufficient condition that show that P is not self-overlapping. tells us when a suitable convex polygon exists for a star For each edge ek we consider the axis-parallel square graph. A e with side length 2k that has vk+1 as a mid- k ⊇ k 0 Concerning the more general caterpillar graphs we point of one of its sides. By our choice of edge lengths provided a basic method for constructing all suitable we observe that Ak 1 Ak and Ak 1 and Ak share a − ⊆ − polygons and we proved that the number of suitable common vertex. We say that a polygon edge belongs to polygons is finite. Furthermore, we showed that an ei if ei is contained in its straight-skeleton face. Let s n/3 2 n-vertex caterpillar graph may possess 2 − suitable denote a polygon edge that belongs to ek+1. By con- polygons. Finding a tight upper bound on the number struction, s cannot overlap with polygon edges belong- of suitable polygons is an open question. Finally, the ing to ek. Using an induction-type argument, all poly- major open questions concern arbitrary trees: How to gon edges belonging to ei, with i < k, are contained in decide feasibility? Are there at most finitely many fea- the one axis-parallel half of Ak that does not contain k+1 sible polygons or is there even an entire continuum of v0 . Hence, s does not intersect polygon edges that feasible polygons? After all, a partial result is given in belong to e1, . . . , ek. It follows by induction that P is [2]: there are at most 2n 5 suitable convex polygons not self-overlapping. for an arbitrary abstract geometric− tree with n leaves.

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