Towards a characterization of stretchable aligned graphs
TTowards a characterization of stretchable alignedgraphs (cid:63)
Marcel Radermacher , Ignaz Rutter , and Peter Stumpf Department of Informatics, Karlsruhe Institute of Technology (KIT), Germany Faculty of Computer Science and Mathematics, University of Passau, Germany [email protected] , {rutter,stumpf}@fim.uni-passau.de Abstract.
We consider the problem of stretching pseudolines in a planarstraight-line drawing to straight lines while preserving the straightnessand the combinatorial embedding of the drawing. We answer open ques-tions by Mchedlidze et al. [9] by showing that not all instances with twopseudolines are stretchable. On the positive side, for k ≥ pseudolinesintersecting in a single point, we prove that in case that some edge-pseudoline intersection-patterns are forbidden, all instances are stretch-able. For intersection-free pseudoline arrangements we show that everyaligned graph has an aligned drawing. This considerably reduces the gapbetween stretchable and non-stretchable instances. Every planar graph G = ( V, E ) has a straight-line drawing [8,11]. In a restrictedsetting one seeks a drawing of G that obeys given constraints, e.g., Biedl etal. [1,2] studied whether a bipartite planar graph has a drawing where the twosets of the partitions can be separated by a straight line. Da Lozzo et al. [4]generalized this result and characterized the planar graphs with a partition L ∪ R ∪ S = V of the vertex set that have a planar straight-line drawing such thatthe vertices in L and R lie left and right of a common line l , respectively, andthe vertices in S lie on l ; refer to Fig. 1a. In this case S is called collinear . Inparticular, they showed that S is collinear if and only if there is a drawing of G such that there is an open simple curve P that starts and ends in the outerface of G , separates L from R , collects all vertices in S and that either entirelycontains or intersects at most once each edge. We refer to P as a pseudoline withrespect to G .Dujmovic et al. [5] proved the following surprising result: If S is a collinearset, then for every point set P with | S | = | P | there is a straight-line drawing Γ of G such that S is mapped to P . Another recent research stream considersthe problem of drawing all vertices on as few lines as possible [3]. Eppstein [7]proved that for every integer l there is a cubic planar graph graph G with O ( l ) vertices such that not all vertices of G can lie on l lines. (cid:63) Work was partially supported by grant RU 1903/3-1 of the German Research Foun-dation(DFG). a r X i v : . [ c s . C G ] A ug Marcel Radermacher, Ignaz Rutter, and Peter Stumpf(a) (b) (c) (d)
Fig. 1: (a) An aligned graph on one (blue) pseudoline. The color indicates thevertex partition L ∪ R ∪ S . (b) Aligned graph of alignment complexity ( ⊥ , , ⊥ ) that does not have an aligned drawing [9]. (c) Allowed types of edges in alignedgraphs of alignment complexity (1 , , . The green edge is aligned. The purpleedge is free. (d) Aligned graph of alignment complexity (2 , , ⊥ ) .Mchedlidze et al. [9] generalized the concept of a single pseudoline with re-spect to an embedded graph to an arrangements of pseudolines and introducedthe notion of aligned graphs , i.e, a pair ( G, A ) where G is a planar embeddedgraph and A = {L , . . . , L k } is a set of pseudolines L i with respect to G thatintersect pairwise at most once. We cite the original definition of aligned draw-ings [9]. A tuple ( Γ, A ) of a (straight-line) drawing Γ of G and line arrangement A is an aligned drawing of ( G, A ) if and only if the arrangement of the unionof Γ and A has same combinatorial properties as the union of G and A . In thefollowing, we specify these combinatorial properties. Let A = { L , L , . . . , L k } ,i.e., line L i corresponds to pseudoline L i . A (pseudo)-line arrangement dividesthe plane into a set of cells C , C , . . . , C (cid:96) . If A is homeomorphic to A , then thereis a bijection φ between the cells of A and the cells of A . If ( Γ, A ) is an aligneddrawing of ( G, A ), then it has the following properties: (i) the arrangement of A is homeomorphic to the arrangement of A , (ii) Γ is a straight-line drawing home-omorphic to the planar embedding of G , (iii) the intersection of each vertex v and each edge e with a cell C of A is non-empty if and only if the intersectionof v and e with φ ( C ) in ( Γ, A ) , respectively, is non-empty, (iv) if an edge uv (directed from u to v ) intersects a sequence of cells C , C , . . . , C r in this order,then uv intersects in ( Γ, A ) the cells φ ( C ) , φ ( C ) , . . . , φ ( C r ) in this order, and(v) each line L i intersects in Γ the same vertices and edges as L i in G , and itdoes so in the same order.Mchedlidze et al. observed that not every aligned graph has an aligned draw-ing. For example, the modification of the Pappus configuration in Fig. 1b doesnot have an aligned drawing. Note that one endpoint of the edge is anchored onsome pseudolines and that the edge crosses three pseudolines. Hence, Mchedldizeet al. studied a restricted subclass of aligned graphs that only contains edges uv that are either (see Fig. 1c and Fig. 1d) – free , i.e, the entire edge uv is in a single cell, – aligned , i.e., the entire edge uv is on a single pseudoline, – one-sided anchored , i.e., u or v is on a pseudoline but not both, and uv doesnot cross a pseudoline, owards a characterization of stretchable aligned graphs 3 – , i.e., u and v are in the interior of a cell and uv crosses one pseu-doline.For this restricted class Mchedlidze et al. proved that every aligned graphhas an aligned drawing. For this purpose they reduced their instances to alignedgraphs that do neither have free edges nor aligned edges nor separating triangles.Then the original instance has an aligned drawing if the reduced instance hasan aligned drawing. Thus, the key to success is to characterize the reducedinstances and to prove that every reduced instance has an aligned drawing.In the reduced setting, Mchedlidze et al. were able to show that each cell ofthe pseudoline arrangement contains at-most a single vertex. Since the unionof two adjacent cells in the line arrangement is convex, any placement of thevertices that respects the ordering constraints along the lines induces a validaligned drawing of the reduced aligned graph. If we additionally allow two-sidedanchored edges , i.e., edges where both endpoints are on pseudolines but that donot cross a pseudoline, then it is possible to construct a family of aligned graphssuch that each cell can contain a number of vertices that is not bounded by thenumber of pseudolines. Contribution.
We show that every aligned graph on k ≥ pseudolines inter-secting in a single point with free, aligned, one-sided and two-sided anchored,and 1-crossed edges has an aligned drawing. If we allow an additional edge type,we show that there is an aligned graph on two pseudolines that does not havean aligned drawing. Note that in the example given in Fig. 1b, no point in thegreen cell is visible from the red vertex within the polygon defined by unionof the (colored) cells traversed by the edge. Hence, this instance trivially doesnot admit an aligned drawing. In contrast, each edge in Fig. 3a can be drawnindependently as a straight-line segment. We show that the entire instance doesnot admit a straight-line drawing. Further, we show that every aligned graph ( G, A ) has an aligned drawing, if A does not have crossings, i.e., A correspondsto an arrangement A of parallel lines. This couples aligned graphs to hierarchi-cal (level) graphs. This significantly narrows the gap in the characterization ofrealizable and non-realizable aligned graphs. We first introduce some notation used for aligned graphs on k pseudolines in-tersecting in a single point. Let O be a point called the origin . Let X = {X , X , . . . , X k } be a pseudoline arrangement where the pseudolines pairwiseintersect in O ; refer to Fig. 2. We refer to an aligned graph ( G, X ) as a k -staraligned graph . Correspondingly, we refer to ( Γ, X ) , with X = { X , X , . . . , X k } as an aligned drawing of ( G, X ) , where the lines in X pairwise intersect in the origin O . The curves in X divide the plane into a set of cells Q , . . . , Q k in coun-terclockwise order. These cells naturally correspond to the regions Q , . . . , Q k bounded by the lines in X .We refer to an edge (vertex) as free if it is entirely in the interior of a cell.An aligned edge (vertex) is entirely on a pseudoline. For each l -crossed edge e Marcel Radermacher, Ignaz Rutter, and Peter Stumpf O Q i X i +1 X i (a) O Q i X i +1 X i (b) Fig. 2: (a,b) (Pesudo)-line arrangements of a -star aligned graph. The greenregion indicates a cell. w w w w v v v v u u u u (a) u v xy λ | x | | y | (cid:27) λ α u β (cid:27) (cid:27) (b) Fig. 3: (a) A -aligned graph that does not have an aligned drawing. (b) We have λ /λ = tan( α ) < tan( β ) = | y | / ( λ + | x | ) .there are l but not l + 1 pseudolines that intersect e in its interior. An edge e is i -anchored if i of its endpoints lie on i distinct pseudolines. Mchedlidzeet al. used a triple ( l , l , l ) , with l i ∈ N ∪ {⊥} to describe the complexity ofan aligned graph ( G, A ) . Let E i be the set of i -anchored edges; note that, theset of edges is the disjoint union E ·∪ E ·∪ E . A non-empty edge set A ⊂ E is l -crossed if l is the smallest number such that every edge in A is at most l -crossed. An aligned graph ( G, A ) has alignment complexity ( l , l , l ) , if E i is at most l i -crossed or has to be empty, if l i = ⊥ . In particular, Mchedlidzeet al. proved that every aligned graph of alignment complexity (1 , , ⊥ ) has analigned drawing. Our results can be restated as that every -star aligned graphof alignment complexity (1 , , has an aligned drawing. Further, there is analigned graph of alignment complexity ( ⊥ , , ⊥ ) that does not have an aligneddrawing. In this section, we study whether k -star aligned graphs have aligned drawings.We first prove that the -star aligned graph in Fig. 3a does not have an aligneddrawing. owards a characterization of stretchable aligned graphs 5(a) (b) (c) e f (d) Fig. 4: (a) This -aligned graph does not have an aligned drawing. (b,c) Thegreen curve indicates the Jordan curve that completes the black edge. The edgein (b) is an edge of a ccw-aligned graph. The edge depicted in (c) is forbiddenin ccw-aligned graphs. (d) A comb of edges e, f . Theorem 1.
There is a -star aligned graph of alignment complexity ( ⊥ , , ⊥ ) that does not have an aligned drawing.Proof. Assume that the aligned graph in Fig. 3a has an aligned drawing. For i =1 , . . . , , with , let ( x i , y i ) be the point for v i , let λ i be the distance of u i to the origin O . Since u v intersects the y -axis above u , edge u v has a steeperslope than the segment u u ; see Fig. 3b. We obtain λ /λ < | y | / ( λ + | x | ) and therefore | x | < λ /λ · | y | . Analogously, we obtain | x i | < λ i +1 λ i · | y i | , i = 1 , | y i | < λ i +1 λ i · | x i | , i = 2 , . (1)Since v i +1 w i are embedded as straight lines, we further obtain estimation (2)that | y i | < | y i +1 | for i = 1 , and | x i | < | x i +1 | for i = 2 , . By multiplying the leftand the right sides we obtain | x | · | y | · | x | · | y | (1) < | y | · | x | · | y | · | x | · λ λ λ λ λ λ λ λ = | y | · | x | · | y | · | x | (2) < | y | · | x | · | y | · | x | . A contradiction. We now consider aligned drawings of k -star aligned graphs ( G, A ) for k ≥ .Recall that the aligned graph in Figure 4a does not have an aligned drawing. Thecrux is that the source of the red edges are free and the source of green edges arealigned. In the following we introduce so-called counterclockwise aligned graphs and show that they have aligned drawings.We orient each non-aligned edge uv of an aligned graph ( G, X ) such thatit can be extended to a Jordan curve, i.e., a closed simple curve, C uv with theproperty that it intersects each pseudoline exactly twice and has the origin toits left. A counterclockwise aligned (ccw-aligned) graph is a k -star aligned graphof alignment complexity (1 , , whose orientation does not contain -anchored -crossed edges with a free source vertex.We prove that every ccw-aligned graph has an aligned drawing. To prove thisstatement we follow the same proof strategy as Mchedlidze et al. In particular, Marcel Radermacher, Ignaz Rutter, and Peter Stumpf we have to augment our aligned graph to a particular ccw-aligned triangulation.Further, we use that for each aligned graph ( G, X ) there is a reduced alignedgraph ( G R , X ) (i.e., it does neither contain (i) separating triangles, nor (ii) freeedges, nor (iii) aligned edges that are not incident to the origin O ) with theproperty that ( G, X ) has an aligned drawing if ( G R , X ) has an aligned drawing;see Lemma 2. In contrast to aligned graphs of alignment complexity (1 , , ⊥ ) the size of ( G R , X ) is not bounded by a constant. The aim of Lemma 3 andLemma 4 is to describe the structure of the reduced instances. This helps toprove Lemma 5 that states that each reduced instance has an aligned drawing.We first introduce further notations. A k -star aligned graph ( G, X ) is a proper k -star aligned triangulation if each inner face is a triangle, the boundary of theouter face is a k -cycle of -anchored edges, the outer face does not contain theorigin and there is a degree- k vertex o on the origin incident to k aligned edges.We refer to a reduced proper ccw-aligned triangulation as a reduced aligned tri-angulation . We refer to -anchored -crossed and -anchored edges as separating .The region within a cell that is bounded by two separating edges e and f is an edge region (Fig. 4d). An inclusion-minimal edge region is a comb .The following lemma is a consequence from the results by Mchedlitze etal. [9]. For further details we refer to the Appendix. Lemma 2.
Every k -star aligned graph has an aligned drawing, if every reduced k -star aligned triangulation has an aligned drawing. Hence, our main contribution is to characterize reduced k -star aligned trian-gulations and then, to prove that every such instance has an aligned drawing. Lemma 3.
Let ( G R , X ) be a reduced aligned triangulation and let o be the vertexon the origin. Then in ( G R − o, X ) each pseudoline X i alternately intersectsvertices and edges, and each comb contains at most one vertex.Proof. Assume that there are two consecutive aligned vertices u and v . Since G R is triangulated and u and v are consecutive, G R contains the edge uv . Thiscontradicts the assumption that ( G R , X ) does not contain aligned edges.The following modification helps us to prove that there are no two consecutiveedges along a pseudoline and that no comb contains two free vertices. u ef Q i ρ i (a) ef Q i ρ i u (b) Fig. 5: The curve ρ i (a) and its modification in (b). owards a characterization of stretchable aligned graphs 7 Let ρ i be the parts of X i and X i +1 that are on the boundary of the cell Q i ,see Figure 5. We modify ρ i as follows. We first, join the endpoints of ρ i in theinfinity such that it becomes a simple closed curve. Let u be a vertex that lieson ρ i . We reroute ρ i such that u now lies outside of ρ i . Since G R is triangulatedand ρ i only intersects edges, ρ i corresponds to a cycle in G (cid:63)R and therefore toa cut C i in G R . Note, each edge of a connected component in G − C i is a freeedge.Now assume that there are two distinct edges e, f that consecutively cross apseudoline X i ∈ X . By the premises of the lemma there is a vertex that lies onthe origin O . Hence both e and f cross X i on the same side with respect to O .Since e and f are distinct and ( G R , X ) is ccw-aligned, there is a cell Q j such that Q j contains two distinct vertices u and w incident to e and f , respectively. Since G is triangulated and e and f are consecutive along X i , u and w are vertices inthe same connected component of G − C j . Therefore, ( G R , X ) contains a freeedge. A contradiction.Consider a comb C in a cell Q i that contains two distinct vertices u and v in its interior. Since G is triangulated and C is inclusion-minimal (it does notcontain another edge-region), u and v belong to the same connected componentof G R − C i . Therefore ( G R , X ) contains a free edge.We call a comb closed if its two separating edges have the same source vertex. Lemma 4.
For every reduced aligned triangulation ( G R , X ) there is a reducedaligned triangulation ( G (cid:48)(cid:48) R , X ) where no closed comb contains a vertex such that ( G R , X ) has an aligned drawing if ( G (cid:48)(cid:48) R , X ) has an aligned drawing.Proof. By Lemma 3 we know that each comb contains at most one vertex. Weapply induction over the number of closed combs that contain a vertex. Let v be a free vertex in a closed comb with separating edges uw , uw . Then weobtain an aligned graph ( G (cid:48) R , X ) by contracting edge uv in the embedding. Since ( G R , X ) is reduced ccw-aligned, all edges outgoing from the free vertex v are -anchored -crossed or -anchored -crossed. In ( G (cid:48) R , X ) they are now 2-anchored0-crossed or 1-anchored 1-crossed with free target vertex. Since there is no othervertex in the comb and the comb is closed, v only has uv as incoming edgewhich is contracted. Therefore ( G (cid:48) R , X ) is ccw-aligned. Assume that ( G (cid:48) R , X ) has an aligned drawing. Since v is a free vertex, we obtain an aligned drawingof ( G, X ) by placing v close to u within in its closed comb. By Lemma 13 weobtain a reduced aligned triangulation ( G (cid:48)(cid:48) R , X ) from ( G (cid:48) , X ) such that ( G (cid:48) R , X ) has an aligned drawing if ( G (cid:48)(cid:48) R , X ) has an aligned drawing. In the constructionthe number of closed combs that contain a vertex is not increased.We can now show that each reduced instance has an aligned drawing. Lemma 5.
Every reduced ccw-aligned triangulation has an aligned drawing.Proof.
By Lemma 4 we can assume that in our triangulation ( G, X ) the closedcombs contain no vertices. By Lemma 3 we know that each comb contains at Marcel Radermacher, Ignaz Rutter, and Peter Stumpf H H H m m m u u vw w o r r (a) v H H m x u (cid:48) u (cid:48) v (cid:48) v (cid:48)(cid:48) u m u r r (b) Fig. 6: (a) Placement of a free vertex v in cell Q . It may be placed within thegray triangle. (b) Example for the observations with u (cid:48) = x and u (cid:48) = x . o H H H H m r r r r m m m Fig. 7: The vertex o and the half-lines H i and the vertices m i , r i for i = 1 , . . . , .All remaining edges and vertices lie in the green area.most one vertex and no vertex if it is closed. The main problem is to draw the -crossed edges. For those, we place each free vertex v close to the right boundaryof its comb. This allows to draw the incoming edges. Since ( G, X ) is ccw-aligned,the target of each -crossed edge vu is free and allows to draw vu .We construct the aligned drawing ( Γ, X ) as follows. Let o be the vertex onthe origin. We call the sources of separating edges corners . First place o and allcorners on X in the order induced from X . For i = 1 , . . . , | X | , let H i be thehalf-pseudoline that is the right boundary of cell Q i . Let m i denote the vertexon H i that is adjacent to o and let r i denote the vertex incident to the outer faceon H i . Note that m i , r i are corners. We write u < i v if u lies between o and v on H i where u , v may be vertices and intersections of edges with H i . Note that < i is a linear order. Define H i correspondingly for X ; see Figure 7. The indicesfor m i , Q i , etc. are considered mod 2 | X | . In the following, we denote by uv theline through two distinct points u , v . Now consider a free vertex v in some cell Q i ; see Figure 6a. It lies in a comb that is bounded by two separating edges u w , u w with u < i u on H i . Note that we have u (cid:54) = u since the combcontains v and is thus not closed. We place v within the triangle bounded by owards a characterization of stretchable aligned graphs 9 m i +1 u , r i +1 u , H i and between m i − u , r i − u (if these lines cross within Q i ,then this means within the triangle bounded by m i − u , r i − u , H i ). Note that v lies in Q i . We will show that the intersections of -crossed edges with H i andthe corners on H i respect the order < i . Finally, we place for i = 1 , . . . , | X | thevertices on H i that are neither o nor a corner arbitrarily on H i respecting theorder < i . This finishes the construction (edges are placed accordingly).We next show that the vertices and edges of G appear for ≤ i ≤ | X | along X i and X i in the same order. Consider the free vertex v and the separating edges u w , u w as defined above. Let m i − = x < i − · · · < i − x k = r i − denotethe corners on H i − . The following three observations imply that all -crossededges with target v cross H i in the correct order between u and u ; refer toFigure 6b.1. m i − v and r i − v cross H i between u and u .2. x v, . . . , x k v intersect H i in the same order as x , . . . , x k lie on H i − .3. Let v (cid:48) be a free vertex in Q i − . Let u (cid:48) w (cid:48) , u (cid:48) w (cid:48) be the separating edges ofthe comb containing v (cid:48) . Then v (cid:48) v crosses H i between u (cid:48) v ∩ H i and u (cid:48) v ∩ H i .For Observation 1, note that v lies between m i − u , r i − u . For Observa-tion 2, note that x v, . . . , x k v cross pairwise in v and thus not in section Q i − .These two observations imply that x v, . . . , x k v cross H i − between u and u .For Observation 3 note now that v (cid:48) lies in the triangle bounded by H i − , u (cid:48) m i and u r i (cid:48) . Observation 3 follows from v and this triangle lying between u m i − and u r i − .We now show that all -crossed edges with target v cross H i in the correctorder between u and u . By Observations 2, 3 the -crossed edges with target v cross H i between m i − v ∩ H i and r i − v ∩ H i . With Observation 1, they cross H i between u and u . By Observation 2, we know that the -anchored -crossededges with target v cross H i in the correct order. By Observations 2, 3, we obtainthat each pair of a 0-anchored 1-crossed and a 1-anchored 1-crossed edge cross H i in the correct order. Since the sources of -anchored -crossed edges withtarget v lie in different combs, they lie pairwise on different sides of some edge x j v by Observation 3. Observation 2 then yields their correct ordering.Since the corners on H i respect < i and all 1-crossed edges have free targetvertices (as the triangulation is ccw-aligned), this implies that the intersectionsof -crossed edges with H i and the corners on H i respect the order < i . Byconstruction, we placed the vertices on H i that are not corners such that theyalso respect order < i . Thus the lines X j intersect the vertices and edges in thesame order as X j .We next show that our embedding is planar by showing that there is nolocation where edges cross. Since the order of intersections with lines in X iscorrect, there are no crossings on X . This leaves us with the cells. Since theseparating edges of Q i appear in the same order on H i and H i +1 , they alsoappear in the same order on H i and H i +1 . Thus, separating edges of the same celldo not cross each other. We further obtain the same combs for ( Γ, XY ) . Consideragain a free vertex v in Q i and the corresponding separating edges u w , u w ; see Figure 6a. Since v lies in the triangle bounded by H i , T and m i +1 u , it alsolies in the comb bounded by u w , u w . Hence, every free vertex lies in thecorrect comb. Let e be an edge incident to v . Then its other end vertex does notlie within the comb of v . It must therefore intersect H i between u and u if itis incoming, and it must intersect H i +1 between u w ∩ H i +1 and u w ∩ H i +1 if it is outgoing. Since we have the same order on H i and H i +1 respectively,edge e crosses neither u w nor u w and thus not the interior of any othercomb in Q i . This means that 1. There are no crossings on separating edges inthe corresponding cells. And that 2. Only edges incident to the free vertex v in a comb intersect the interior of that comb. Those edges are all adjacent in v and do not cross. We obtain that there are no crossings on X , no crossingson separating edges in the corresponding cells and no crossings within combs.Hence, our embedding is planar.Since there are no free edges and the order of intersections with lines in X isfixed, the order of incident edges around a free vertex is also fixed. For a vertex u on X we note that each adjacent free vertex is in another comb and thereforethe order of incident edges around u is also fixed. Therefore, our embedding Γ induces the same combinatorial embedding as the embedding of G .From Lemma 2 and Lemma 5 we directly obtain our main theorem. Theorem 6.
Every ccw-aligned graph ( G, X ) has an aligned drawing. In this section, we prove that every aligned graph ( G, A ) has an aligned drawing,if A is intersection free, i.e., the line arrangement A is a set of parallel lines.Our result uses a result of Eades at al. [6], and of Pach and Toth [10]. Eadeset al. consider hierarchical plane graphs. A graph G = ( V, E ) with a mappingof the vertices to a layer L i is a hierarchical graph , where a set of layers L is aset of ordered parallel horizontal lines L i ∈ L . A hierarchical plane drawing of ahierarchical graph is a planar drawing where each vertex is on its desired layerand each edge is drawn as a y -monotone curve. Two hierarchical drawings are equivalent if each layer, directed from −∞ to ∞ , crosses the same set of edgesand vertices in the same order. Eades et. al. [6] proved that for every hierarchicalplanar drawing of a graph there is an equivalent hierarchical planar straight-linedrawing. Pach and Toth [10] proved a similar result stating that for every y -monotone drawing where no two vertices have the same y -coordinate there isan equivalent y -monotone straight-line drawing such that each vertex keeps its y -coordinate. In contrast to these two results, we have that the y -coordinate isonly prescribed for a subset of the vertices, i.e., there are some (free) verticesthat have to be positioned between two layers (lines). The proof strategy is toextend the initial pseudoline arrangement with an additional set of intersection-free pseudolines such that there are no free vertices.Due to [9] (compare Lemma 2), we can assume that there are neither freenor aligned edges. For the purpose of this section, a reduced aligned graph is an owards a characterization of stretchable aligned graphs 11 aligned graph that has no aligned edges and no free vertices. Note that previouslyonly free edges were forbidden. Thus, the current definition is more restrictive.The following theorem is an immediate corollary from the results of Eades etal. [6], and Pach and Toth [10]. Theorem 7.
For every intersection-free pseudoline arrangement, every reducedaligned graph ( G, A ) has an aligned drawing. Lemma 8.
Let A be an intersection-free pseudoline arrangement and let A bea line arrangement homeomorphic to A . For every aligned graph ( G, A ) thereis a reduced aligned graph ( G, A (cid:48) ) such that A ⊂ A (cid:48) and ( G, A ) has an aligneddrawing if ( G, A (cid:48) ) has an aligned drawing. L E L L v Fig. 8: Construction of the new pseudoline L v (red) that contains v . The red-dotted pseudoline L (cid:48) v indicates the copy of L (bottom blue) that crossed theedges in E L (green) in the same order as L Proof.
We first insert for each free vertex v a new pseudoline L v to A such that v is on L . Thus, the aligned graph ( G, A (cid:48) ) does not have free vertices.Let L be a pseudoline that is on the boundary the region R v of A thatcontains v . Let E L be the set of edges of G that are (partially) routed through R v and that are either crossed by L or that have an endpoint on L . We assumethat L is directed. Then the direction of L induces a total order of the edges in E L . We obtain a curve L (cid:48) v that crosses the edges in E L in this order and in theirinterior. Since v is free, G is triangulated and ( G, A ) contains neither free noraligned edges, there is at-least one edge e ∈ E L that is incident to v . Denote by e f and e l in E L the first and last edge incident to v . We obtain a pseudoline L v that contains v from L (cid:48) v by rerouting L (cid:48) v along e f and e l such that it is does notcross these edges in their interior and such that v is on the line (Fig. 8).Now, let ( G, A (cid:48) ) be the aligned graph that is obtained by the previous pro-cedure for each free vertex v . Let A (cid:48) be any set of parallel lines that contains A and corresponds to A (cid:48) . Clearly, ( Γ, A ) is an aligned drawing of ( G, A ) if ( Γ, A (cid:48) ) is an aligned drawing of ( G, A (cid:48) ) . This finishes the proof.Theorem 7 and Lemma 8 together prove the following theorem. Theorem 9.
Let A be an intersection-free pseudoline arrangement and let A be a (parallel) line arrangement homeomorphic to A . Then every aligned graph ( G, A ) has an aligned drawing ( G, A ) . In the paper, we showed that every aligned graph ( G, A ) has an aligned draw-ing if ( G, A ) is either a ccw-aligned graph or if A is intersection-free. Further,we provided a non-trivial example of a -star aligned graph that does not ad-mit an aligned drawing. Thus, in our opinion the most intriguing open ques-tion is whether every aligned graph of alignment complexity (1 , , has analigned drawing, for general stretchable pseudoline arrangements A . Our ex-ample shows that this statement is not true for aligned graphs of alignmentcomplexity (1 , , . Our stretchability proof of counterclockwise aligned graphsuses the fact that we can move each free vertex v to an aligned vertex u on thecell of v . Performing this operation for all free vertices at once ensures that wedo not introduce edges of a forbidden alignment complexity. Figure 9 indicatesthat for general aligned graphs of alignment complexity (1 , , there is not al-ways a consistent mapping of free vertices to aligned vertices such that that theresulting graph has the same alignment complexity. Thus it is unclear whetherthe techniques used in the paper can be used to decide whether every alignedgraph of alignment complexity (1 , , has an aligned drawing. (a) (b) Fig. 9: There is no mapping of free vertices to aligned vertices on the boundaryof the same cell such that moving the free vertices onto their image results in analigned graph of alignment complexity (1 , , . References
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A Reducing k -star aligned graphs In this section, we give further details on how to reduce a k -star aligned graph.We first recall the definition of proper and reduced triangulations. A k -staraligned graph ( G, X ) is a proper k -aligned triangulation if each inner face isa triangle, the boundary of the outer face is a k -cycle of -anchored edges, theouter face does not contain the origin and there is a degree- k vertex o on theorigin incident to k aligned edges. We refer to a reduced proper ccw-alignedtriangulation as a reduced aligned triangulation if it does neither contain (i) sep-arating triangles, nor (ii) free edges, nor (iii) aligned edges that are not incidentto the origin O .)Mchedldize et al. proved the following triangulation lemma. Lemma 10.
For every aligned graph ( G, X ) of alignment complexity (1 , , ⊥ ) there is an aligned triangulation ( G (cid:48) , X ) of alignment complexity (1 , , ⊥ ) suchthat G is a subgraph of G (cid:48) . Fig. 10: (a) Illustration of the key properties of a proper k -star alignedgraph.(b) Examples of allowed (black) edges in a reduced instance and forbidden(red) edges. w w w (cid:48) u (cid:48) vu v (cid:48) w (cid:48) (a) (b) (c) Fig. 11: (a) The (black) separating edges are isolated by the green edges. (b) Theblack edges are removed and the red edges are obtained by the triangulation.(c) Final graph, after removing edges in the interior of a quadrangle u, w , v, w and reinserting the black edges. owards a characterization of stretchable aligned graphs 15 Since ccw-aligned graphs contain -anchored and -anchored -crossed edges,we can not immediately apply this lemma. In the following, we show that our in-stances can be modified such that the can use the previous lemma. For simplicity,we assume that there is no aligned that crosses the origin. Lemma 11.
Let ( G, X ) be a ccw-aligned graph. Then there is a ccw-alignedtriangulation ( G (cid:48) , X ) that contains ( G, X ) as a subgraph. Moreover, the outerface of ( G (cid:48) , X ) is bounded by k -cycle C of -anchored edges and the outer facedoes not contain the origin in its interior.Proof. Let ( G , X ) be the graph that is constructed from ( G, X ) as follows. First,add a k -cycle C of -anchored edges in the outer face such that the new outerface does not contain the origin.For each separating edge uv of G add two vertices w , w and the edges uw , w v and uw , w v . Route and direct the edges according to Figure 11a. Fi-nally, remove the edge uv . Eventually, we arrive at an aligned graph of alignmentcomplexity (1 , , ⊥ ) . With the application of Lemma 10 we obtain a triangulatedaligned graph ( G , X ) of alignment complexity (1 , , ⊥ ) . We remove edges in theinterior of each quadrangle u, w , v, w and reinserted the original edge uv . Fi-nally, we remove all edges and vertices in the region bounded by C that doesnot contain the origin. This yields the desired aligned graph ( G (cid:48) , X ) .Since no free edge of an ccw-aligned graph is incident to a triangle thatcontains the intersection in its interior, the next lemma follows from the resultsof Mchedlitze et al. Lemma 12.
Let ( G, X ) be a ccw-aligned graph and let e be an interior free edgeor an aligned edge that is neither an edge of a separating nor a chord and doesnot contains the origin, then ( G/e, X ) is a ccw-aligned graph and ( G, X ) has analigned drawing if ( G/e, X ) has an aligned drawing. Thus, we can now prove the main reduction lemma and therefore Lemma 2.
Lemma 13.
For every ccw-aligned graph ( G, X ) there is a reduced aligned tri-angulation ( G R , X ) such that ( G, X ) has an aligned drawing if ( G R , X ) has analigned drawing.Proof. By Lemma 11 there is a aligned triangulation ( G T , X ) of ( G, X ) withthe outer face bounded by k -cycle of -anchored edges. Moreover, an aligneddrawing of ( G T , X ) contains an aligned drawing of ( G, X ) .By Mchedlidze et al. we obtain a reduced aligned triangulation ( G (cid:48) R , X ) from ( G T , X ) by either splitting ( G T , X ) into two aligned graphs at a separatingtriangle T , or by contracting free or aligned edges that are not incident to o (Lemma 12). Moreover, we have that that ( G T , X ) has an aligned drawing if ( G (cid:48) R , X ) has an aligned drawingIn order to obtain a proper aligned triangulation ( G R , X ) from ( G (cid:48) R , X ) weperform the reduction depicted in Figure 12. If there is an aligned edge thatcontain the origin in its interior, we place a subdivision vertex on this edge ( G (cid:48) R , X Y ) :( G R , X Y ) : (a) vuv v vu u uvu vu (b)
Fig. 12: Red edges are removed from ( G T , X ) and green added to ( G P , X ) and inserted edges as depicted in Figure 12a. Note that in this case an aligneddrawing of ( G R , X ) contains an aligned drawing of ( G (cid:48) R , X ) .Consider the case that there is a vertex v on the origin that is incident to a freevertex u . We obtain a new aligned graph ( G R , X ) by exhaustively applying thereductions depicted in Figure 12b. Since the black polygon (compare Figure 12b)in an aligned drawing of ( G R , X ) is star-shaped and its kernel contains thevertex v , ( G (cid:48) R , X ) has an aligned drawing if ( G (cid:48) R , X ))