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Linear size universal point sets for classes of planar graphs⋆
Stefan Felsner[0000−0002−6150−1998], Hendrik Schrezenmaier[0000−0002−1671−9314], Felix Schr¨oder[0000−0001−8563−3517], and Raphael Steiner[0000−0002−4234−6136]
Institute of Mathematics, Technische Universit¨atBerlin, Germany {felsner,fschroed,schrezen,steiner}@math.tu-berlin.de
2 1 Abstract. A finite point set P ⊆ R is n-universal with respect to a 2 class G if every n-vertex-graph G ∈ G admits a crossing-free straight-line 3 drawing with vertices being placed at points of P . A popular task in 4 graph drawing is to identify small universal point sets. 5 For the class of all planar graphs the best known upper bound on the 6 size of a universal point set is quadratic and the best known lower bound 7 is linear in n. One of the classical results in the area is that every set 8 of n points in general position (no three collinear) is n-universal for out- 9 erplanar graphs. While some other classes are known to admit universal 10 point sets of near linear size, we are not aware of truly linear bounds for 11 interesting classes beyond outerplanar graphs. 12 In this paper we study a specific ordered point set H (the exploding 13 double chain) and show that all planar graphs G on n ≥ 2 vertices which 14 are subgraphs of a planar graph admitting a one-sided Hamiltonian cycle 15 have a straight-line drawing on the initial piece Hn of size 2n − 2 in H. ′ 16 Let H be the class of all subgraphs of planar graphs admitting a one- 17 sided Hamiltonian cycle. It had been conjectured that all 4-connected ′ 18 triangulations belong to H . While the conjecture has been disproved, it 19 is still true that Hn is n-universal for a large class H of planar graphs. We 20 show that all bipartite plane graphs and all cubic plane graphs belong ′ ′ 21 to H ⊆ H. To our surprise, however, not all planar 2-trees are in H .
Keywords: universal point set · one-sided Hamiltonian · 2-book em- beddings.
22 1 Introduction
2 23 Given a family G of planar graphs and a positive integer n, a point set P ⊆ R is 24 called an n-universal point set for the class G or simply n-universal for G if for
25 every graph G ∈ G on n vertices there exists a straight-line crossing-free drawing
26 of G such that every vertex of G is placed at a point of P . It is a widely studied
⋆ This research started at the 4th Workshop within the collaborative DACH project Arrangements and Drawings, February 24–28, 2020, in Malchow. Supported by the German Research Foundation DFG Project FE 340/12-1. 2 S. Felsner et al.
27 and fundamental open problem in geometric graph theory (compare also the
28 entry [14] in the Open Problem Garden) to determine, given a class of graphs G, 29 (the asymptotics of) the minimum size fG(n) of an n-universal point set for G. 30 If G is the class of all planar graphs we simply write f(n) := fG(n).
31 Schnyder [18] showed that for n ≥ 3 the regular (n − 2) × (n − 2)-grid forms
32 an n-universal point set for planar graphs, even if the combinatorial embedding 2 2 33 of the planar graph is prescribed. This shows that f(n) < n = O(n ). Asymp-
34 totically, the quadratic upper bound on f(n) remains the state of the art up
35 until today. However, the multiplicative constant in this bound was improved 1 2 36 in follow-up papers, see [4,5]. The current upper bound is f(n) ≤ 4 n + O(n) 37 by Bannister et al. [4]. For several subclasses G of planar graphs, better upper
38 bounds are known: A classical result by Pach et al. [16] is that every outerplanar
39 n-vertex graph embeds straight-line on any set of n points in general position, 40 and hence fout-pl(n) = n. Near-linear upper bounds of fG(n) = O(n polylog(n)) 41 are known for 2-outerplanar graphs, simply nested graphs, and for the classes
42 of bounded pathwidth [3,4]. Finally, for the class G of planar 3-trees (also
43 known as apollonian networks or stacked triangulations), an upper bound of 3/2 44 fG(n) = O(n log n) has been proved by Fulek and T´oth[12].
45 As for lower bounds, the trivial bounds n ≤ fG(n) ≤ f(n) hold for all n ∈ N 46 and all planar graph classes G. The best lower bound f(n) ≥ 1.293n − o(n)
47 from [17] has been shown using planar 3-trees, we refer to [6,13,8,9] for earlier
48 work on lower bounds.
49 It seems that in order to improve the quadratic upper bound on f(n) to 2 50 o(n ), the considered point sets should be not too uniformly distributed. Indeed,
51 Choi, Chrobak and Costello [7] recently proved that point sets chosen uniformly 2 52 at random from the unit square must have size Ω(n ) in order to be universal
53 for n-vertex planar graphs, with high probability.
54 In this paper we study a specific ordered point set H (the exploding double 55 chain) and let Hn be the initial piece of size 2n−2 in H (for n ≥ 2). Throughout 56 the paper, denote by H be the class of all planar graphs G which have a plane 57 straight line drawing on the point set Hn with n = |V (G)|. That is, Hn forms 58 an n-universal point set for H.
59 A graph is a POSH (partial one-sided Hamiltonian) if it is a spanning sub-
60 graph of a planar graph admitting a one-sided Hamiltonian cycle. Our main ′ 61 result (Theorem 1) is that every POSH is in H. We let H = {G : G is POSH}.
′ 62 Theorem 1 motivates the study of H . This class of planar graphs seems
63 to be quite large, e.g., the smallest 4-connected triangulation which is known
64 not to be POSH has 113 vertices [2]. On the positive side we show that every
65 bipartite plane graph is POSH (Section 3). In Section 4 we use the construction
66 for bipartite graphs to show that cubic plane graphs are POSH. Section 5 has a
67 negative result, namely that not all 2-trees are POSH. We conclude with some
68 conjectures and open problems. Linear size universal point sets 3
69 2 The point set and the embedding strategy
′ 70 In this section we introduce an ordered point point set H and a class H of 71 planar graphs and show that for every n ≥ 2 the initial part Hn of size 2n − 2 ′ 72 is n-universal for the class H . 73 A sequence Y = (yi)i≥0 of real numbers satisfying y1 = 0, y2 = 0, and√ 74 yi+1 > 2yi + yi−1 for all i ≥ 2 is called exploding. Note that if α > 1 + 2, i−3 75 then y1 = y2 = 0 and yi = α for i ≥ 3 is an exploding sequence. Given an 76 exploding sequence Y let P (Y ) = (pi)i≥0 be the set of points with pi = (i, yi) ¯ 77 and let P (Y ) = (qi)i≥0 be the set of points with qi = (i, −yi), i.e., the reflected ¯ 78 point set, and note that p1 = q1 and p2 = q2. Let H(Y ) = P (Y ) ∪ P (Y ) and 79 Hn(Y ) = {pi, qi|1 ≤ i ≤ n} so that |Hn(Y )| = 2n−2. Figure 1 illustrates H6(Y ). Let H = H(Y ) for some exploding sequence Y . For two points p and q let H(p, q) be the set of points of H in the open right half-plane of the directed line −→pq. Note that1 (p ) ∪ (p ) ∪ (q ) if i > j k k≤j k k>i ℓ ℓ
83 Moreover, if i < j then H(qi, qj) = H(pi, qj) \{qi} and if i > j then H(pi, pj) = 84 H(pi, qj) \{pj}. These sidedness informations characterize the order type of H. 85 A point set A = {pi, qi|i ≥ 1} is an exploding double chain if it has the order 86 type of H.
p6 q6
p5 q5
p4 q4
p3 q3 x
p2 q2 y p1 q1
87 Fig. 1. An example of a point set H6 in a rotated coordinate system.
88 A plane graph G has a one-sided Hamiltonian cycle with reverse edge vu if 89 it has a Hamiltonian cycle (v1, v2, . . . , vn) such that u = v1, v = vn and uv is 90 incident to the outer face. Moreover, for every j = 2, . . . , n in the restriction 91 G[v1, . . . , vj, vj+1] of G the edges vj−1vj and vj+1vj are consecutive in the rota- 92 tion of vj. A more pictural reformulation of the second condition is that in the
1 80 In cases where i or j are in {1, 2} the following may list one of the two points defining 81 the halfspace with its second name as member of the halfspace. For correctness such 82 listings have to be ignored. 4 S. Felsner et al.
93 embedding of G for every j either all the back-edges vivj with i < j are drawn 94 inside the closed bounded region D whose boundary is the the Hamiltonian cycle 95 or they are all drawn in the closed outside of the cycle. We let VI be the set 96 of vertices vj which have a back-edge vivj with i < j − 1 drawn inside D and 97 VO = V \ VI . 98 In the context of cartograms Alam et al. [1] conjectured that every plane
99 4-connected triangulation has a one-sided Hamiltonian cycle. Later Alam and
100 Kobourov [2] found a plane 4-connected triangulation on 113 vertices which has
101 no one-sided Hamiltonian cycle. ′ 102 Recall that H is the class of POSH graphs, i.e., of plane graphs which are
103 spanning subgraphs of plane graphs admitting a one-sided Hamiltonian cycle.
104 Our interest in this class is motivated by the following theorem.
′ 105 Theorem 1. Let G be a POSH and let v1, . . . , vn be a one-sided Hamiltonian ′ 106 cycle of a plane supergraph G of G on the same vertex set. Then there is a ′ 107 crossing-free embedding of G on Hn with the property that vi is placed on either 108 pi or qi.
109 Proof. It is sufficient to describe the embedding of the supergraph G on Hn. 110 For the proof we assume that in the plane drawing of G the sequence v1, . . . , vn 111 traverses the boundary of D in counter-clockwise direction. For each i vertex vi 112 is embedded atv ¯i = pi if vi ∈ VI and atv ¯i = qi if vi ∈ VO. (If the boundary 113 of D is traversed in clockwise direction we exchange the assignment. The proof
114 then follows from the counter-clockwise case by considering the mirror of G, its 115 embedding on Hn and then the mirror of the embedding.) 116 Let Gi = G[v1, . . . , vi] be the subgraph of G induced by {v1, . . . , vi}. The 117 path Λi = v1, . . . , vi separates Gi, the left part GLi consists of the intersection 118 of Gi with D, the right part GRi is Gi minus all edges which are interior to 119 D. The intersection of GLi and GRi is Λi and their union is Gi. The counter- 120 clockwise boundary walk of Gi consists of a path ∂Ri from v1 to vi which is 121 contained in GRi and a path from vi to v1 which is contained in GLi, let ∂Li 122 be the reverse of this path. ¯ 123 Let Gi be the straight line drawing of the plane graph Gi induced by placing ¯ 124 each vertex vj at the correspondingv ¯j. A vertexv ¯ of Gi is said to see a point p ¯ 125 if there is no crossing between the segmentvp ¯ and an edge of Gi. 126 With induction we show that for every i ∈ {2, . . . , n}: ¯ 127 1. The drawing of Gi is plane, i.e., non-crossing. ¯ 128 2. Gi and Gi have the same outer boundary walks. ¯ 129 3. Every vertex of ∂Li in Gi sees all the points pj with j > i and every vertex ¯ 130 of ∂Ri in Gi sees all the points qj with j > i.
131 For i = 2 the graph Gi is just an edge and the three claims are immediate, 132 for property 3 just recall that the line spanned by p1 and p2 separates the p-side 133 and the q-side of Hn. ¯ 134 Now assume that i ∈ {3, . . . , n}, the properties are true for Gi−1 and suppose 135 that vi ∈ VI (the argument in the case vi ∈ VO works symmetrically). This Linear size universal point sets 5
136 implies that all the back-edges of vi are in the interior of D whence all the 137 neighbors of vi belong to ∂Li−1. Since vi ∈ VI we havev ¯i = pi and property 3 ¯ ¯ 138 of Gi−1 implies that the edges connecting tov ¯i can be added to Gi−1 without ¯ 139 introducing a crossing. This is property 1 of Gi. ¯ 140 Since Gi−1 and Gi−1 have the same boundary walks and vi respectivelyv ¯i ¯ 141 belong to the outer faces of Gi and Gi and since vi has the same incident edges ¯ ¯ 142 in Gi asv ¯i in Gi, the outer walks of Gi and Gi again equal each other, i.e., 143 property 2. 144 Let j be minimal such that vjvi is an edge and note that ∂Li is obtained 145 by taking the prefix of ∂Li−1 whose last vertex is vj and append vi. The line ¯ 146 spanned byv ¯j andv ¯i = pi separates all the edges incident tov ¯i in Gi from 147 all the segmentsv ¯ℓpk with ℓ < j andv ¯ℓ ∈ ∂Li and k > i. This shows that ¯ 148 every vertex of ∂Li in Gi sees all the points pk with k > i. For the proof of the 149 second part of property 3 we refer to Figure 2, it shows that the new edgesv ¯jpi 150 do not obstruct the visibility between vertices of ∂Ri and any qk with k > i. 151 Of course this can also be derived formally by translating the condition for a
152 crossing between two segments into sidedness conditions and then compare with
153 the sidedness conditions given for the order type of H. This completes the proof
154 of property 3. 155 We have shown the three properties for the case where vi ∈ VI . The other 156 case vi ∈ VO is symmetric. Finally, property 1 for Gn implies the theorem. ⊓⊔
pi qk
qj qk
157 Fig. 2. Vertices from ∂Ri see qk
158 3 Plane bipartite graphs
159 In this section we consider bipartite planar graphs and show that they are POSH.
160 Theorem 2. Every bipartite plane graph G = (V,E) is a subgraph of a plane ′ 161 graph G on the same vertex set V which has a one-sided Hamiltonian cycle,
162 i.e., G is POSH.
163 Quadrangulations are the plane graphs with all faces of degree four. Equiva-
164 lently they are the maximal plane bipartite graphs, i.e., any edge-addition either 6 S. Felsner et al.
165 breaks bipartiteness or planarity. Every bipartite plane graph is a spanning sub-
166 graph of a plane quadrangulation. Since every spanning subgraph of a POSH is
167 a POSH it suffices to prove the theorem for plane quadrangulations. 168 Let Q be a quadrangulation and let VB and VW be the color classes of a 169 2-coloring. Label the two black vertices of the outer face as s and t. Henceforth,
170 when talking about a quadrangulation we think of an embedded quadrangulation
171 endowed with s and t.A separating decomposition is a pair D = (Q, Y ) where
172 Q is a quadrangulation and Y is an orientation and coloring of the edges of Q
173 with colors red and blue such that:
174 1. The edges incident to s and t are incoming in color red and blue, respectively. 175 2. Every vertex v ̸∈ {s, t} is incident to a non-empty interval of red edges and
176 a non-empty interval of blue edges. If v is white, then, in clockwise order,
177 the first edge in the interval of a color is outgoing and all the otheredges
178 of the interval are incoming. If v is black, the outgoing edge is the clockwise
179 last in its color (see Figure 3).
180 Fig. 3. Edge orientations and colors at white and black vertices.
181 Separating decompositions of a quadrangulation Q have been defined by
182 de Fraysseix and Ossona de Mendez [15]. They show a bijection between separat-
183 ing decompositions and 2-orientations (orientations of the edges of Q such that
184 every vertex v ̸∈ {s, t} has out-degree 2) and show the existence of a 2-orientation
185 of Q with an argument related to flows and matchings. An inductive proof for
186 the existence of separating decompositions was given by Felsner et al. [11], this
187 proof is based on identifying pairs of opposite vertices on faces.
188 It is known that in a separating decomposition the red edges form a tree
189 directed towards s, and the blue edges form a tree directed towards t. Each of
190 the trees connects all the vertices v ̸∈ {s, t} to the respective root. Proofs for
191 this can be found in either of [15,11,10]. The later two of these sources continue
192 to show that the edges of the two trees can be separated by a curve which starts
193 in s, ends in t, and traverses every vertex and every inner face of Q. This curve
194 is called the equatorial line.
195 If Q is redrawn such that the equatorial line is mapped to the x-axis with s
196 being the left end and t being the right end of the line, then the red tree and
197 the blue tree become alternating trees (defined below) drawn in the upper re-
198 spectively lower half-plane defined by the x-axis. Note that such a drawing of Q Linear size universal point sets 7
199 is a 2-book embedding, we call it an alternating 2-book embedding to emphasize
200 that the graphs drawn on the two pages of the book are alternating trees.
t
s t
s
201 Fig. 4. A quadrangulation Q with a separating decomposition S, and the alter-
202 nating 2-book embedding induced by the equatorial line of S.
203 An alternating tree is a plane tree T with a plane drawing such that the ver-
204 tices of T are placed at different points of the x-axis and all edges are embedded
205 in the half-plane above the x-axis (or all below). Moreover, for every vertex v
206 it holds that all its neighbors are on one side, either they are all left of v or all
207 right of v. In these cases we call the vertex v respectively a right or a left vertex
208 of the alternating layout.
209 Let Q be a plane quadrangulation on n vertices and let S be a separating 210 decomposition of Q. Let s = v1, v2, . . . , vn = t be the spine of the alternating 2- + 211 book embedding of Q based on S. Let Q be obtained from Q by adding vnv1 and 212 all the edges vivi+1 which do not yet belong to the edge set of Q. By construction + 213 v1, v2, . . . , vn is a Hamiltonian cycle of Q and due to the alternating property of 214 the embedding this Hamiltonian cycle is one-sided with reverse edge vnv1 = ts. 215 Hence Q is POSH.
216 It is worth noting that the Hamiltonian cycle read in the reverse direction, 217 i.e., as vn, vn−1, . . . , v1, is again one-sided, now the reverse edge is v1vn = st.
218 4 Plane cubic graphs
219 In this section we identify another large subclass of the class H and more precisely ′ 220 of the subclass H of graphs which are POSH.
221 Theorem 3. Every plane cubic graph G is a spanning subgraph of a plane ′ 222 graph G on the same vertex set V which has a one-sided Hamiltonian cycle,
223 i.e., G is POSH. 8 S. Felsner et al.
224 We will use Theorem 2, therefore, we need to associate a plane quadrangu-
225 lation with G. To this end, we find a matching whose contraction results ina
226 bipartite graph.
227 Lemma 1. Let G be a cubic graph. Then G admits a matching M such that
228 contracting all the edges of M results in in a bipartite multi-graph.
229 Proof. Pick a partition X ∪ Y = V (G) of the vertex-set of G such that the
230 size of the cut, i.e., the number of edges in G with exactly on endpoint in X
231 and exactly one endpoint in Y , is maximized. We claim that G[X] and G[Y ]
232 are graphs of maximum degree at most 1 (and thus, a disjoint union of isolated
233 vertices and a matching). Suppose that a vertex v ∈ X has at least two neighbors
234 in G[X]. Then v has at most one neighbor in Y , and hence moving v from X
235 to Y increases the size of the cut by at least one, a contradiction. The same
236 argument works to show that G[Y ] has maximum degree at most one.
237 Let M be the matching in G consisting of all the edges in G[X] or G[Y ].
238 Contracting all the edges in M transforms G[X] and G[Y ] into independent
239 sets, and hence results in a bipartite multi-graph G/M. ⊓⊔
240 Given a cubic plane graph G we find a matching M as in Lemma 1 and let B
241 be the plane bipartite multi-graph obtained from G by contracting the edges ′ 242 in M. Let B be the underlying simple graph of B and let Q be a quadrangulation ′ 243 which has B as a spanning subgraph. The proof of Theorem 2 shows that there 244 is a left to right placement v1, . . . , vn of the vertices of Q on the x-axis such that 245 for each i ∈ [n] all the edges vjvi with j < i − 1 are in one half-plane and all 246 edges vivj with j > i + 1 are in the other half-plane. Delete all the edges from Q ′ 247 which do not belong to B , and duplicate the double-edges of B in the drawing.
248 This yields an embedding Γ of B.
249 Vertices of degree four in B are exactly the edges which have been obtained
250 by contracting edges of M. We now show how to undo the contractions, i.e.,
251 split vertices of degree four, in the drawing Γ in such a way that at the end we ⋆ 252 arrive at a one-sided 2-book drawing Γ of G, that is, a 2-book embedding of G 253 with vertex-sequence v1, . . . , vn such that for every j ∈ {1, . . . , n} the incident 254 back-edges vivj with 1 ≤ i < j are all drawn either on the spine or on the same 255 page of the book embedding (all above or all below the spine). Once we have
256 obtained such a book embedding, we can add the spine edges to G to obtain a + 257 supergraph G of G which has a one-sided Hamiltonian path. This then shows
258 that G is POSH.
259 Let v be a vertex of degree four of B. Vertex v was obtained by contracting 260 an edge uw ∈ M. Label the edges of v in clockwise order as e1, e2, e3, e4 such 261 that in G the edges e1, e2 are incident to u and e3, e4 are incident to w. If the 262 two angles ∠e2e3 and ∠e4e1 together take part of both half-planes defined by 263 the x-axis, then it is possible to select two points left and right of the point 264 representing v in Γ and to slightly detour the edges ei such that no crossings are 265 introduced and one of the two points is incident to e1, e2 and the other to e3, e4. 266 Addition of an edge connecting the two points completes the split of v into the
267 edge uw ∈ M. Figure 5 shows a few instances of this local split. Linear size universal point sets 9
e2 e2 e1 e1 e1 e1 e4 e4 e4 e4 v u e3 e3 w v e2 e2 w u e e 3 3 e e2 e3 e2 e3 2 e2 e e 1 1 e3 e v w v u 3 u w e4 e4 e1 e1 e4 e4
268 Fig. 5. Four cases for the local split of a vertex v.
269 The above condition about the two angles is not fulfilled if all four edges
270 of v emanate into the same halfspace and the clockwise numbering starting at 271 the x-axis is either e4, e1, e2, e3 or e2, e3, e4, e1. The two cases are the same up 272 to exchanging the names of u and w, therefore, we can assume the first one. 273 Let v1, v2, v3, v4 be the neighbors of v in left to right order. A more important 274 distinction is whether all vi are left of v or to the right of v. In the first case we 275 leave w at the former position of v and u slightly left of v3 while in the second 276 case u is slightly right of v2. Figure 6 shows the first case and Figure 7the 277 second. To see that in the first case edges uv2 and uw are free of crossings we 278 can route them close to the path v3vv2 and the edge v3v in the original drawing. 279 This kind of split is a far split.
wv v1 v2 v3 v4 v1 v2 u v3 v4
280 Fig. 6. A far split of a vertex v with edges to the left in the upper half-plane.
281 Changes are local to the gray region.
284 A special case occurs when a vertex v of degree four has a double-edge. This
285 either leads to a local split (a few more cases for the complete listing of all
286 possible local splits) or the double-edge and two simple edges are in the same 287 halfplane such that v1 and v4 are the vertices incident to the simple edges, 288 whence, v2 = v3. We stick to the rules of the far split. If v has edges to the 289 left we place u to the left of v3 and connect to v2 and v3 this makes a short 290 double-edge on the spine. The case where edges from v go to the right is again
291 symmetric. 10 S. Felsner et al.
v w v1 v2 v3 v4 v1 v2 u v3 v4
282 Fig. 7. A far split of a vertex v with edges to the right in the upper half-plane.
283 Changes are local to the gray region.
292 By now we have shown that starting from the drawing Γ of B we can split
293 a single vertex v of degree four into an edge uw ∈ M. To split all the degree
294 four vertices we proceed as follows. First we split all vertices which are subject
295 to a far split. Let U be the set of vertices of edges uw ∈ M which are far, i.e.,
296 w stays at the position of the contraction vertex v and u is the vertex close to 297 either v2 or v3. 298 We claim that every vertex has at most one neighbor in U to its left and one ′ 299 to its right. Indeed if a vertex v has a neighbor of U to its right (see Figure 7), ′ ′ 300 then v was the v2 of some far split and v was the leftmost neighbor of v . The ′ 301 claim now follows from the observation that v only has one leftmost neighbor. ′ 302 It can happen that a vertex v which was classified as being subject of a local 303 split looses this property and has to undergo a far split. For example vertex v2 304 in Figure 7 has the neighbor v on the left but may have three neighbors to the 305 right, the split of v generates a fourth neighbor on the right of v2. When no 306 further far split is possible we do all the local splits. These local splits can not
307 conflict with each other. This completes the proof of Theorem 3.
308 5 2-Trees
309 From the positive results in the previous two sections it could be expected that
310 essentially every “sufficiently sparse” planar graph is POSH. In this section we
311 show that this is not true. 312 A 2-tree is a graph which can be obtained, starting from a K2, by repeatedly 313 selecting an edge of the current graph and adding a new vertex which is made
314 adjacent to the endpoints of that edge. We refer to this operation as stacking a
315 vertex over an edge.
316 From the recursive construction it follows that a 2-tree on n vertices is a
317 planar graph with 2n − 3 edges. We also mention that 2-trees are series-parallel
318 planar graphs. Another well studied class which contains 2-trees as a subclass is
319 the class of planar Laman graphs.
320 Fulek and T´othhave shown that planar 3-trees admit n-universal point sets 3/2 321 of size O(n log n). Since every 2-tree is an induced subgraph of a planar 3-tree
322 the bound carries over to this class.
323 Proposition 1. There is a 2-tree G on 499 vertices which is not POSH. Linear size universal point sets 11
324 Throughout the following discussion we assume that a 2-tree G is given to- 325 gether with a a left to right placement v1, . . . , vn of the vertices on the x-axis 326 such that adding the spine edges and the reverse edge vnv1 to G we obtain a 327 plane graph with an one-sided Hamiltonian cycle.
328 For an edge e of G we let X(e) be the set of vertices which are stacked over e
329 and S(e) the set of edges which have been created by stacking over e, i.e., each
330 edge in S(e) has one vertex on e and one vertex in X(e). We partition the set 331 X(e) of an edge e = vivj with i < j into a left part XL(e) = {vk ∈ X(e): k < i}, 332 a middle part XM(e) = {vk ∈ X(e): i < k < j}, and a right part: XR(e) = 333 {vk ∈ X(e): j < k}.
334 Claim 1 For every edge |XR(e)| ≤ 2.
335 Suppose that |XR(e)| ≥ 3. Each vertex in this set has all its back-edges on the
336 same side. Two of them use the same side for the back edges to the vertices of e.
337 This implies a crossing pair of edges, a contradiction.
′ ′ 338 Claim 2 If for all e ∈ S(e) we have |X(e )| ≥ 3, then |XM(e)| ≤ 3.
339 Suppose that e = vivj with i < j is in the upper half-plane and there are 340 four vertices x1, x2, x3, x4 in XM(e). One-sidednes implies that the four edges 341 xkvj are in the upper half-plane. Now if x1, x2, x3, x4 is the left to right order, 342 then the edges vix2, vix3, and vix4 have to be in the lower half-plane. Now let ′ ′ 343 e = vix3 and consider the three vertices in X(e ). Two of them, say y1, y2, are 344 on the same side of x3. ′ 345 First suppose y1, y2 ∈ X(e ) are left of x3. Note that the edges of vix2 and 346 x2vj enforce that y1, y2 are between x2 and x3. Due to the edge x2vj the edges 347 viy1 and viy2 are in the lower half-plane. One-sidedness at x3 requires that y1x3 348 and y2x3 are also in the lower half-plane. This makes a crossing unavoidable. ′ 349 Now suppose that y1, y2 ∈ X(e ) are right of x3. The edges vix4 and x4vj 350 enforce that y1, y2 are between x3 and x4. Due to the edge x3vj the edges viy1 351 and viy2 are in the lower half-plane. Now let y1 be left of y2. One-sidedness at y2 352 requires that x3y2 is also in the lower half-plane, whence, there is a crossing 353 between viy1 and x3y2. This completes the proof of the claim.
′ 355 Claim 3 If XL(e) ≥ 2 and x is the rightmost element of XL(e), then XL(e ) ≤ 2 ′ 356 for some e ∈ S(e) incident with x.
357 Suppose that e = vivj with i < j is in the upper half-plane and there are 358 two vertices x1, x2 in XL(e). We assume that x2 is the rightmost element of 359 XL(e). From one-sidedness at vj we know that x1vj and x2vj are in the upper 360 half-plane. Now x1vi and hence also x2vi are in the lower half-plane. All the 361 vertices of X(x2vi) and X(x2vj) are in the region bounded by x2vj, vjvi, vix2. 362 Suppose that we have y1, y2 ∈ XL(x2vi) and z1, z2 ∈ XL(x2vj). By one-sidedness 363 the edges from x2 to the four vertices y1, y2, z1, z2 are in the same half-plane. If 364 they are in the lower half-plane and y1 is left of y2 there is a crossing between 365 y1x2 and y2vi. If they are in the upper half-plane and z1 is left of z2 there is a 12 S. Felsner et al.
Claim 2 e Claim 3
vi x3 x4 vj x1 x2 e x1 x2 vi vj
e′
354 Fig. 8. Illustrating the proofs of the claims.
′ 366 crossing between z1x2 and z2vj. The contradiction shows that XL(x2vi) ≤ 2 or 367 XL(x2vj) ≤ 2, since x = x2 this completes the proof of the claim.
368 We are ready to define the graph G and then use the claims to verify that G
369 is not a POSH. The graph G contains a base edge e, and there are seven vertices
370 stacked on e, i.e., |X(e)| = 7. Edges in S(e) are the primary edges of G. For ′ ′ 371 each primary edge e there are five vertices stacked on e . Edges introduced by
372 stacking over a primary edge are secondary edges. Finally, for each secondary ′′ ′′ 373 edge e there are three vertices stacked on e . Note that there are 7 · 2 = 14
374 primary edges, 14 · 5 · 2 = 140 secondary edges and 140 · 3 · 2 = 840 edges
375 introduced by stacking on a secondary edge. In total the number of edges is
376 995 = 2n − 3, hence, the graph has 499 vertices. 377 Now suppose that G is POSH and let v1, . . . , vn the order of vertices on the 378 spine of a certifying 2-book embedding. Let vivj with i < j be the vertices of 379 the base edge e. By symmetry we may assume that e is in the upper half-plane.
380 From Claim 1 we get |XR(e)| ≤ 2 and from Claim 2 we get |XM(e)| ≤ 3, it 381 follows that |XL(e)| ≥ 2. Let x1 and x2 be elements of XL(e) such that x2 is ′ ′′ 382 the rightmost element of XL(e). Let e = x2vi and e = x2vj and observe that ′ ′′ 383 XR(e ) = ∅ = XR(e ) (the argument can be found in the proof of Claim 3). From ′ ′′ ′ ′′ 384 Claim 2 applied to e and e we deduce that |XM(e )| ≤ 3 and |XM(e )| ≤ 3. ′ ′′ 385 Hence |XL(e )| ≥ 2 and |XL(e )| ≥ 2. This is in contradiction with Claim 3.
386 We conclude that there is no spine ordering for G which leads to a one-sided
387 crossing free 2-book embedding.
388 6 Concluding remarks
389 We have introduced the exploding double chain as a special point set (respec- 390 tively order type) which has the property that the initial segment Hn of size 391 2n − 2 is n-universal for graphs on n vertices which are POSH, i.e., which are
392 subgraphs of plane graphs admiting a one-sided Hamiltonian cycle. We have
393 shown that all bipartite plane graphs and all plane cubic graphs are POSH. In
394 fact we believe that the class is much larger and propose two conjectures in this
395 direction: Linear size universal point sets 13
396 Conjecture 1. Every triangle-free plane graph is POSH.
397 Conjecture 2. Every 5-connected planar triangulation is POSH.
398 We have shown that 2-trees and their superclasses series-parallel and planar ′ 399 Laman graphs are not contained in the class H of POSH graphs. The question
400 whether these classes admit universal point sets of linear size remains intriguing.
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