DDrawing outer-1-planar graphs revisited Therese Biedl David R. Cheriton School of Computer Science, University of Waterloo,Waterloo, Ontario N2L 1A2, Canada.
Abstract
In a recent article (Auer et al, Algorithmica 2016) it was claimed thatevery outer-1-planar graph has a planar visibility representation of area O ( n log n ). In this paper, we show that this is wrong: There are outer-1-planar graphs that require Ω( n ) area in any planar drawing. Then wegive a construction (using crossings, but preserving a given outer-1-planarembedding) that results in an orthogonal box-drawing with O ( n log n )area and at most two bends per edge. A graph is a graph that can be drawn in the plane such that every edgehas at most one crossing. Many graph-theoretic and graph-drawing results areknown for 1-planar graphs, see for example [12]. One subclass of 1-planar graphsis the class of outer-1-planar (o1p) graphs , which have a 1-planar drawing suchthat additionally every vertex is on the outer-face (the unbounded region of thedrawing).Outer-1-planar graphs were introduced by Eggleton [9] and studied by manyother researchers [1, 2, 7, 11]. Of particular interest to us is a paper byAuer, Bachmeier, Brandenburg, Gleißner, Hanauer, Neuwirth and Reislhuber[2]. Among others, they characterize the forbidden minors of outer-1-planargraphs, give a recognition algorithm, and give bounds on various graph param-eters such as number of edges, treewidth, stack number and queue number. Fi-nally they turn to drawing algorithms for outer-1-planar graphs, and here claimthe following result: “Every o1p graph has a planar visibility representation in O ( n log n ) area.” (Theorem 8).In this paper, we show that this result is incorrect. Specifically, we constructan n -vertex outer-1-planar graph such that in any planar embedding there areΩ( n ) nested triangles (we give detailed definitions below). It is known [10] thatany planar graph drawing with k nested cycles requires width and height at least2 k in any planar poly-line drawing. Since any planar visibility representationcan be converted into a poly-line drawing of asymptotically the same width andheight [5], any planar visibility representation of our graph uses Ω( n ) area andthe claim by Auer et al. is incorrect. Research supported by NSERC. The author would like to thank Jayson Lynch for pointingout that the lower bound also holds for IC-planar graphs. a r X i v : . [ c s . C G ] S e p hen we give drawing algorithms for outer-1-planar graphs that do achievearea o ( n ). These drawings have crossings, but we can reflect exactly the givenouter-1-planar embedding. Our construction gives orthogonal box-drawingswith area O ( n log n ) and at most two bends per edge; they can be converted topoly-line drawings of the same area.To our knowledge, the only prior result on orthogonal drawings of outer-1-planar drawings (other than the one by Auer et al. that we disprove) is byArgyriou et al. [1]; they showed that every outer-1-planar graph with maximumdegree 4 has an outer-1-plane point-orthogonal drawing with O ( n ) area and atmost 2 bends per edge. We assume familiarity with graphs, see e.g. [8]. A planar graph is a graph thatcan be drawn in the plane without any crossing. Such a drawing Γ defines the regions , which are the connected parts of R \ Γ. The infinite region is calledthe outer-face . A planar drawing defines the planar embedding consisting of the rotation scheme (the clockwise order of edges at each vertex) and the outer-face (a lists of vertices and edges on the outer-face). A graph is called outer-planar if it has a planar embedding where all vertices are on the outer-face.A is a graph that can be drawn in the plane such that everyedge has at most one crossing. As above one defines regions and outer-face ofsuch a drawing. An outer-1-planar graph is a graph with a 1-planar drawingwhere additionally all vertices are on the outer-face. Any such drawing is de-scribed via an outer-1-planar embedding , consisting of the rotation scheme, theouter-face, and information as to which pair of edges cross.In this paper we almost only consider maximal outer-planar and maximalouter-1-planar graphs, which are those graphs where as many edges as possiblehave been added while staying in the same graph class and having no duplicateedges or loops.A poly-line drawing of a graph is a drawing where vertices are points andedges are polygonal curves; a bend is the transition-point between segments ofthe polygonal curve. We also consider orthogonal box-drawings , where verticesare represented by axis-aligned boxes and edges are polygonal curves with onlyhorizontal and vertical segments. A special kind of orthogonal box-drawing is a visibility representation where edges have no bends.The orthogonal box-drawings created in this paper are somewhat specializedin that vertices are flat : All vertex-boxes are actually horizontal line segments(in the figures, we show them thickened into a thin rectangle). We call such avertex-box a bar and such an orthogonal box-drawing an orthogonal bar-drawing .We assume (without further mentioning) that all our drawings are grid-drawings , i.e., all defining features (vertex-points, endpoints of vertex-bars,bends) are placed at points with integer coordinates. We measure the width and height of a grid-drawing as the number of vertical/horizontal grid-lines thatintersect the smallest enclosing bounding box of the drawing. We call a drawing2 rder-preserving if it exactly reflects a given (planar or 1-planar) embedding ofthe graph.
In this section, we construct an outer-1-planar graph that requires Ω( n ) areain any planar poly-line drawing. Our graph G L (for L ≥ × L -grid with every second region filled with a crossing. Clearly this isan outer-1-planar graph, see Figure 1. Enumerate the vertices of G L as in thefigure. v v w w v L w L v L − w L − v L w L x y G L \ K T
Figure 1: The outer-1-planar graph G , and how to find nested triangles.It is known that all outer-1-planar graphs have a planar drawing [2], butthey can have many different planar embeddings. However, we can show thatfor our graph G L , all planar embeddings are bad in some sense.Call a set of disjoint triangles T , . . . , T (cid:96) nested (in a fixed planar embedding)if for i = 2 , . . . , (cid:96) the region bounded by T i includes all vertices of T ∪ · · · ∪ T i − . Lemma 1
Fix L ≥ even. Any planar embedding Γ of G L with ( v L , w L ) onthe outer-face contains L/ nested triangles. Proof:
Set K := { v L , w L , v L − , w L − } . These four vertices form a K ; itsinduced embedding Γ K is hence unique up to renaming. By assumption theouter-face T of Γ K contains v L , w L and one vertex y ∈ { v L − , w L − } ; set x = { v L − , w L − } \ y .If L = 2, then we are done (use triangle T ). If L >
2, then graph G (cid:48) := G L \ K is connected, so must reside entirely within one face f of Γ K . Graph G (cid:48) containsneighbours of x and y , so face f must contain both x and y . Since x is not onthe outer-face of Γ K , face f is not the outer-face of Γ K . So no vertex of G L \ K is in the outer-face of Γ K , making T the outer-face of the entire drawing Γ.Observe that G (cid:48) = G L \ K is a copy of G L − . Since both v L − and w L − haveneighbours in { x, y } , edge ( v L − , w L − ) is on the outer-face of the induced draw-ing Γ (cid:48) of G (cid:48) . By induction, Γ (cid:48) contains L/ − T , . . . , T L/ − .Adding the outer-face T to this gives the desired set of nested triangles for G since G L − resides inside T and is vertex-disjoint from it. (cid:3) heorem 1 There exists an n -vertex outer-1-planar graph that requires widthand height at least n/ in any planar poly-line grid-drawing. Proof:
Fix an arbitrary integer N , and consider graph G N which has n = 8 N vertices. Observe that G N contains two disjoint copies of G N , obtained byremoving the edges ( v N , v N +1 ) and ( w N , w N +1 ). In any planar embeddingof G N , at least one of these two copies of G N must have its rightmost/leftmostgrid-edge on the outer-face of its induced planar embedding. In this copy, wetherefore have N nested triangles by Lemma 1. It is known [10] that (cid:96) nestedtriangles require width and height 2 (cid:96) in any planar poly-line drawing, whichimplies the result by 2 N = n/ (cid:3) If we use so-called [4], then the lower bound can beimproved ever-so-slightly to ( n + 2) / all innerregions of the 2 × L -grid; see a preliminary version of this paper [6] for details.We gave the weaker bound here because G L has two other advantages: it is IC-planar (no two crossings have a common vertex) and it has maximum degree4 (so the lower bound even holds for orthogonal point-drawings).
We now briefly review the algorithm by Auer et al. [2] to explain where the errorlies. Their algorithm is based on a prior algorithm (we call it here
MaxOutpl )by the author that creates an order-preserving orthogonal bar-drawing of anymaximal outer-planar graph [3, 4]. The idea
MaxOutpl is to fix one reference-edge ( s, t ) with poles s, t on the outer-face (with s before t in clockwise order).Then draw graph G such that the bars of s and t occupy the top right andbottom right corner respectively.To do so, split the graph and recurse, see also Figure 2(a). Specifically,consider the interior face incident to ( s, t ) (say its third vertex is x ). Of thetwo subgraphs “hanging” at the edges ( s, x ) and ( x, t ), pick the smaller one.(Formally, for any edge ( u, v ) (cid:54) = ( s, t ), the hanging subgraph H uv is the graphinduced by all outer-face vertices between v and u , using the path from v to u that does not include edge ( s, t ).) Assume that H x,t is not bigger than H s,x ,but has at least three vertices (all other cases are handled symmetrically orwith another simpler construction). Let { x, y , t } be the other interior faceat ( x, t ). Recursively draw the three subgraphs H s,x , H x,y and H y ,t withrespect to reference-edges ( s, x ) , ( x, y ) and ( y , t ). After a minor modificationof the drawings (“releasing” one pole, defined below) and rotating the drawingof H x,y , these drawings can be merged as shown in Figure 2(b).Auer et al. [2] used the same idea, but release other poles, mirror drawingsrather than rotate them, and route edge ( x, t ) differently. This leaves spacefree to also route edge ( s, y ) and removes all bends, hence giving a visibilityrepresentation. See Figure 2(c). However, there are a few issues with thisconstruction: 4 txH s,x H y ,t y H x,y (a) H s,x H (cid:48) x,y H (cid:48) y ,t tsx y (b) tsx y H s,x H (cid:48) x,y H (cid:48) y ,t (c) H s,x H (cid:48) x,y H (cid:48) y ,t tsx y (d) Figure 2: The construction by the author [4] and by Auer et al. [2]. (a) Splittingthe graph (the red dashed edge does not exist for outer-planar graphs). (b)Putting drawings together in [4]. (c) Putting drawings together in [2]. (d) Avariation of [2]. • First, the logarithmic height-bound for
MaxOutpl crucially requires thatthe constructed drawing is no bigger than the drawing of the bigger sub-graph H s,x . This is violated in the construction from Figure 2(c), thoughthe issue can easily be fixed by drawing one edge horizontally instead, seeFigure 2(d). • Second, Auer et al. silently assume that the region incident to ( s, t ) hasa crossing. If it does not, but if the other region incident to ( x, t ) has acrossing, then it is not even clear how y should be picked, and the crossingedges are not both drawn. • Finally, even if the region at ( s, t ) has a crossing, the crossing edge neednot be ( s, y ). (Recall that in MaxOutpl vertex y is determined bythe size of the subgraphs and cannot be picked arbitrarily.) Instead, thecrossing edge could connect t to a vertex in H s,x , and we cannot add suchan edge to the drawing without adding bends or going through other bars.The third issue is the one that led to our counter-example, constructed suchthat if we pick { s, t, x, y } to be the endpoints of a crossing, then graph H s,x is not the biggest of the subgraph (and neither is H y ,t ), and so the logarithmicheight-bound fails to hold. 5 Constructions
It should be quite obvious that if we allow crossings and some bends, then we can create drawings of area O ( n log n ) for any outer-1-planar graph G . Specifically,pick an arbitrary maximal outer-planar subgraph G − of G , and let Γ − be anorthogonal bar-drawing of G − obtained with MaxOutpl . Since every edge hasat most two bends, every region of Γ − has O (1) bends. As such, any edge of G \ G − (which needs to be drawn through two adjacent regions) can be insertedwith O (1) bends. We now work on reducing this bound on the number of bendsand show: Theorem 2
Any outer-1-planar graph has an order-preserving orthogonal box-drawing with at most two bends per edge and O ( n log n ) area. It is straight-forward to convert any planar orthogonal box-drawing into apoly-line drawing while keeping the area asymptotically the same and adding atmost two bends per edge. See [5] for details, and note that the same techniqueworks whether the drawing is planar or not. Therefore our result implies:
Corollary 3
Any outer-1-planar graph has an order-preserving poly-line draw-ing with at most four bends per edge and O ( n log n ) area. Since we have a constant number of bends per edge, and any outer-1-planargraph has O ( n ) edges [2], we have O ( n ) vertical segments in the orthogonalbox-drawing. As such, after deleting empty columns if needed, the width isautomatically O ( n ). Therefore all our analysis is focused on the height of thedrawing, which we prove to be in O (log n ). Now we prove Theorem 1 with a recursive drawing algorithm. We roughly followthe idea of
MaxOutpl , but explicitly distinguish whether the region incidentto ( s, t ) is crossed or not. Crucially, we allow more types of drawings for thesubgraphs to achieve fewer bends overall.We only draw maximal outer-1-planar graphs; one can always make an outer-1-planar graph maximal by adding edges, and delete those edges from the ob-tained drawing later. It is known that for a maximal outer-planar graph theedges on the outer-face have no crossing [9].So fix a maximal outer-1-planar graph G with a fixed outer-1-planar em-bedding and with reference-edge ( s, t ). An orthogonal bar-drawing Γ of G iscalled • a drawing of type A if the bars of s and t occupy the top right and bottomright corner of Γ, respectively (this is the same as for [4]); • a drawing of type B if the bar of s occupies the top right corner of Γ, andthe bar of t occupies the point one row below this corner;6 a drawing of type B if the bar of t occupies the bottom right corner of Γ,and the bar of s occupies the point one row above this corner; • a drawing of type C if the bars of s and t occupy the bottom left andbottom right corner of Γ, respectively.All drawings that we create are order-preserving. In particular edge ( s, t ) mustbe drawn clockwise along the boundary of the drawing; Figure 3 shows how itwill be drawn. A st B st B st C s t C s t A = B = B st Figure 3: The drawing-types, and the base cases.Let α ≈ .
59 be such that α = (1 − α ) . Let φ := √ − ≈ .
618 be suchthat φ = 1 − φ . Define γ := max {− φ , − α } ≈ max { . , . } = 3 . , wehence know γ log α ≤ − , γ log(1 − α ) = γ log( α / ) = γ log α ≤ − γ log φ ≤ − , γ log(1 − φ ) = γ log( φ ) = 2 γ log φ ≤ − , Also set δ = 2. We measure the size | G | of an n -vertex maximal outer-1-planar graph G as n −
1; this may be rather unusual but will help keep theequations simpler. Define h ( G ) := γ log | G | + δ ≈ .
94 log( n −
1) + 2; this isthe height that we want to achieve in our drawings. Theorem 2 now holds if weshow the following result.
Lemma 2
Let G be a maximal outer-1-planar graph with reference-edge ( s, t ) .Then G has order-preserving orthogonal bar-drawings with at most two bendsper edge and of the following kind: • A type-A drawing A of height at most h ( G ) , • a type- B drawing B of height at most h ( G ) + 2 , • a type- B drawing B of height at most h ( G ) + 2 , and • a type-C drawing C of height at most h ( G ) + 3 .Furthermore, at least one of B and B has height at most h ( G ) . We prove Lemma 2 by induction on | G | . In the base case, G consists of onlyedge ( s, t ), and one easily constructs suitable drawings, even without bends. SeeFigure 3. The height is at most 2 in all cases. Since | G | = 1, we have log | G | = 0and the bound holds by δ = 2. 7 .2 Subgraphs and tools Now assume that n ≥
3, so G has at least one inner region. The idea is to split G into subgraphs, recursively obtain their drawings, and put them togethersuitably. We have two cases (see also Figure 4). In Case ∆, the inner regionat ( s, t ) has no crossing; by maximality it is hence a triangle, say { s, t, x } . Wewill recurse on the two hanging subgraphs H s,x and H x,t , and use H L := H s,x and H R := H x,t as convenient shortcuts. Observe that | H L | + | H R | = | G | sincewe define the size to be one less than the number of vertices. In Case × theinner region at ( s, t ) is incident to a crossing, say edge ( s, y ) crosses edge ( t, x ).By maximality the edges ( s, x ), ( x, y ) and ( y, t ) exist and have no crossing.We will recurse on the three hanging subgraphs H L := H s,x , H M := H x,y and H R := H y,t . Observe that | H L | + | H M | + | H R | = | G | . These (two or three)subgraphs are smaller, and we assume that they have been drawn inductively,giving drawings A L , B L , B L , C L for subgraph H L , and similarly for the othersubgraphs. In the pictures, we use A L for drawing A L rotated by 180 degrees,and similarly for other drawing-types and subgraphs. s txH L = H s,x H R = H x,t s txH L = H s,x H R = H y,t yH M = H x,y Figure 4: Splitting a subgraph.To put drawings together, we frequently use two well-known tools [3, 4]: • If we have a drawing Γ of some subgraph, then we can insert empty rowsto increase its height since the drawing is orthogonal. If we choose theplace to add empty rows suitably, then this does not change the type ofthe drawing. • If we have a drawing Γ of some subgraph, with vertex s in the top row,then we can release s : add a new row above Γ, let s occupy all of this row,and re-route edges to neighbours of s . (If s has a horizontal neighbour z , then the edge ( s, z ) now becomes vertical.) See Figure 5 and [4] fordetails. This increases the height by 1, and achieves that s now occupiesboth the top left and top right corner in the resulting drawing Γ (cid:48) .Similarly we can release vertex t to occupy the bottom-left corner, pre-suming it is drawn in the bottom row. In the pictures, we use a “prime”(e.g. A (cid:48) L as opposed to A L ) to indicate that one pole has been released; itwill be clear from the picture which one.8igure 5: Releasing vertex s . Picture taken from [4]. × We start with Case × where the region incident to ( s, t ) has a crossing, and studythe three different types of drawings that we want to achieve. We occasionallyuse h := h ( G ) as convenient shortcut. Case × .A: We want a type-A drawing of height h . We distinguish sub-casesby the size of H M . Sub-case × .A.1: | H M | ≤ α | G | (recall that α ≈ . | H L | + | H R | ≤ | G | , hence we may assume | H R | ≤ | G | / × .A. ( a )from Figure 6. (The case | H L | ≤ | G | / × .A. ( b ).) A L A (cid:48) M A (cid:48) R tsx y (a) A R A (cid:48) M A (cid:48) L tsxy (b) B M A (cid:48) L C R tsx y (c) B M A (cid:48) R C L tsxy (d) Figure 6: Constructions for × .A .We will (for this case only) explain in detail how this figure is to be inter-preted; for later cases we hope that the figures alone suffice. We use drawings A L , A M and A R of the subgraphs H L , H M , H R . The primes in the figure indi-cate that we should release release y in both A M and A R to get A (cid:48) M and A (cid:48) R .Rotate A (cid:48) M by 180 ◦ to get A (cid:48) M . Increase the height of drawings, if needed, suchthat A (cid:48) R and A (cid:48) M have the same height; then combine the two bars of y into9ne. Increase the height of A M , if needed, so that it is at least two rows tallerthan the other two drawings. Then we combine these drawings and route theedges ( s, y ) , ( x, t ) and ( s, t ) as shown in Figure 6(a).One can easily verify that the result is an order-preserving drawing. Toargue that it has height at most h , the general procedure is as follows. Firststudy the height of all three drawings of subgraphs, which must be at most h . Furthermore, at some of these subgraph-drawings more rows are needed, forreleasing vertices and/or routing edges and/or other bars. If this is the case, thenwe must argue that the subgraph-drawing is sufficiently much smaller ( h − × .A. h at this subgraph-drawing, and so other parts of the drawing are not forced to increase beyondheight h . After arguing this for all three subgraphs, we therefore know that theheight of the constructed drawing is at most h .In the specific case here, the height-analysis is done as follows. Since | H L | ≤| G | , drawing A L has height at most h ( H L ) ≤ h ( G ) = h . We have | H M | ≤ α | G | ,so drawing A M has height at most h ( H M ) = log | H M | + δ ≤ γ log( α · | G | ) + δ = γ log | G | + δ + γ log α = h ( G ) + γ log α ≤ h − γ log α ≤ −
3. Since | H R | ≤ | G | < α | G | , likewise A R has height at most h −
3. We need three more rows above A M and A R : one to release y , one foredge ( x, t ) and one for the bar of s . So the height requirement is at most h everywhere as desired. Sub-case × .A.2: | H M | > α | G | . We know that one of B M or B M has heightat most h ( G M ). Let us assume that this is B M , and we then use construction × .A. ( c ) from Figure 6 (the other case uses construction × .A. ( d ) and is similarlyanalyzed).Drawing B M has height at most h ( G M ) ≤ h by assumption. Since | H R | ≤| G | − | H M | ≤ (1 − α ) | G | , drawing C R has height at most h ( H R ) + 3 ≤ γ log((1 − α ) | G | ) + δ + 3 = h ( G ) + γ log(1 − α ) + 3 ≤ h − γ log(1 − α ) ≤ −
5. We need two more rows above C R (for ( x, t ) and the barof s ) and the height requirement is at most h here. Drawing A L has height atmost h ( H L ), which by | H L | ≤ (1 − α ) | G | is similarly shown to be at most h − A L (for releasing x and the bar of y ). Therefore theheight requirement is at most h everywhere. Case × .B: We want two drawings, of type B , B . Both have height at most h + 2, and one has height at most h .Consider first constructions × .B. ( a ) for a type- B drawing, and × .B. ( b ) fora type- B drawing, see Figure 7. In both, the drawing of H M has height at most h ( H M ) + 2 ≤ h + 2. Drawings A L and A R have height at most h , and we needtwo more rows at them (one to release a pole, one for a bar of a vertex not inthe subgraph). So either drawing has height at most h + 2 as desired.10 M A (cid:48) L A (cid:48) R x y ts (a) B M A (cid:48) R A (cid:48) L xy ts (b) A L A (cid:48) M B (cid:48) R xx y ts (c) A R A (cid:48) M B (cid:48) L xy ts (d) Figure 7: Constructions for × .B and × .B .But we must distinguish cases (and perhaps use a different construction) toachieve that one of the drawings has height at most h . Sub-case × .B.1: | H L | , | H R | ≤ φ | G | (recall that φ = ( √ − / ≈ . B M or B M has height at most h ( G M ). Let us assume that thisis B M , and consider again construction × .B. ( a ) (in the other case one similarlyanalyzes construction × .B. ( b )).Drawing B M by assumption has height at most h . Also for i ∈ { L, R } wehave | H i | ≤ φ | G | and drawing A i has height at most h ( H i ) = γ log | H i | + δ ≤ γ log | G | + δ + γ log φ ≤ h ( G ) + γ log φ ≤ h − γ log φ ≤ −
2. We need two further rows at each of A L and A R , so theheight requirement is at most h everywhere. Sub-case × .B.2: | H L | > φ | G | . Use construction × .B. ( c ) to obtain a type- B drawing, see Figure 7. We know that | H R | ≤ (1 − φ ) | G | and hence B R hasheight at most h ( H R ) + 2 ≤ h ( G ) + 2 + γ log(1 − φ ) ≤ h − γ log(1 − φ ) ≤ −
4. We need two more rows above it (one to release t andone for the bar of s ), so the height requirement at B R is at most h . Similarly theheight of A M is at most h ( H M ) ≤ h −
4. We need four more rows above it: onerow for releasing y , one row because B (cid:48) R has a row between y and the (released) t , one row for ( x, t ) and one row for the bar of s . So the height requirement isat most h everywhere. Sub-case × .B.3: | H R | > φ | G | . Symmetrically construction × .B. ( d ) gives atype- B drawing of height h . 11 ase × .C: We want a type-C drawing of height h + 3. Sub-case × .C.1: | H M | ≥ | G | . We know that one of H L , H R has size atmost ( | G | − | H M | ) ≤ | G | . Let us assume that this is H L , and we then useconstruction × .C. ( a ) from Figure 8 (the other case uses construction × .C. ( b )and is similarly analyzed). C L A (cid:48) M A (cid:48) R ts x y (a) C R A (cid:48) M A (cid:48) L x y ts (b) A (cid:48) R B (cid:48) M A (cid:48) L x y ts (c) A (cid:48) L B (cid:48) M A (cid:48) R x y ts (d) Figure 8: Constructions for × .C .Drawings A M and A R both have height at most h , and we need three morerows (one to release y , one for ( x, t ) and one for ( s, t )) so the height requirementhere is h + 3. By | H L | ≤ | G | , drawing C L has height at most h ( H L ) + 3 = γ log | H L | + δ + 3 ≤ h ( G ) + γ log( ) + 3 ≤ h − γ ≥
2. We require three more rows above it: one for ( s, y ), one row that wasused for ( x, t ) elsewhere, and one row for ( s, t ). So the height requirement hereis actually strictly less than h + 3. Sub-case × .C.2: | H M | ≤ | G | . We know that one of H L , H R has size at most | G | . Let us assume that this is H R , and we then use construction × .C. ( c ) fromFigure 8 (the other case uses construction × .C. ( d ) and is similarly analyzed).Drawing A L has height at most h , and we need three more rows (for releasing x , edge ( s, y ) and edge ( s, t )), so the height requirement here is at most h + 3.Drawing B M has height at most h ( H M ) + 2 ≤ h ( G ) + γ log( ) + 2 ≤ h by γ ≥
2. Again we need three more rows, so the height requirement is at most h + 3. Similarly by | H R | ≤ | G | drawing A R has height at most h −
2. We needfive more rows at A R : one row for releasing y , one row because B (cid:48) M had one row12etween y and the (released) x , one row for ( x, t ), one row that was used for( s, y ) elsewhere, and one row for ( s, t ). So the height requirement everywhere isat most h + 3. ∆ Now we turn our attention to the (much simpler) case ∆ where the region inci-dent to edge ( s, t ) has no crossing. We again distinguish cases by the drawing-type that we want to achieve.
Case ∆ .A: We want a type-A drawing of height h . Sub-case ∆ .A.1: | H L | , | H R | ≤ φ | G | . Then use construction ∆ .A. ( a ) fromFigure 9. Drawing B L has height at most h ( H L ) + 2 = γ log | H L | + δ + 2 ≤ h ( G ) + γ log φ + 2 ≤ h since γ log φ ≤ −
2. Similarly drawing A R has height at most h − x and the bar of s ). So the heightrequirement is at most h everywhere. B L A (cid:48) R tsx (a) A L C R tsx (b) A R C L tsx (c) Figure 9: Constructions for ∆ .A . Sub-case ∆ .A.2: | H L | > φ | G | . Then use construction ∆ .A. ( b ). Drawing A L has height at most h . By | H R | ≤ | G | − | H L | ≤ (1 − φ ) | G | , drawing C R has heightat most h ( H R ) + 3 ≤ h ( G ) + γ log(1 − φ ) + 3 ≤ h − γ log(1 − φ ) ≤ −
4. We require one more row above it (for the bar of s ),so the height requirement is at most h everywhere. Sub-case ∆ .A.3: | H R | > φ | G | . Symmetrically one proves that construction∆ .A. ( c ) has height at most h . Case ∆ .B: We want two drawings, of type B , B . Both have height at most h + 2, and one has height at most h .The construction for the type- B drawing is in Figure 10(a). Since A L and A R have height at most h , and we need two further rows above A R , the heightrequirement is at most h + 2 everywhere. If | H R | ≤ | G | then the height of A R is at most h ( H R ) ≤ h ( G ) + γ log ≤ h − L A (cid:48) R tsx (a) A R A (cid:48) L tsx (b) A (cid:48) R A (cid:48) L ts x (c) Figure 10: Constructions for ∆ .B and ∆ .B , as well as for ∆ .C .by γ ≥ h everywhere.Likewise, the construction of a type- B drawing in Figure 10(b) has heightat most h + 2, and if | H L | ≤ | G | then the height is at most h . Since one of H L and H R has size at most | G | , one of the drawings has height at most h . Case ∆ .C : We want a type-C drawing of height at most h + 3. The construc-tion is shown in Figure 10(c). Since A L and A R have height at most h , and weneed two more rows above them (to release x and for edge ( s, t )), the height isactually at most h + 2. We have given suitable constructions in all cases, so by induction Lemma 2holds. Using the type-A drawing, we get a drawing of height 3 .
94 log( n −
1) + 2and width O ( n ). Therefore Theorem 2 holds. Following the proof, one also seesthat the drawing can easily be found in linear time, since we can construct thefour drawings of each hanging subgraph in constant time from the drawings ofits subgraph. In this paper, we pointed out an error in a result by Auer et al., and showthat for some outer-1-planar graphs, any poly-line drawing without crossingsrequires Ω( n ) area. We then studied orthogonal box-drawings of outer-1-planargraphs that achieve small area. We create such drawings (using bars to representvertices) that have O ( n log n ) area and at most 2 bends per edge, and exactlyreflect the given outer-1-planar embedding.We believe that reducing the number of bends per edge should be possible,and in particular, conjecture that we can achieve O ( n log n ) area with at mostone bend per edge, perhaps at the expense of modifying the 1-planar embedding.On the other hand, finding drawings with 0 bends (i.e., visibility representations)appears difficult. Perhaps bar-1-visibility drawings (where edges are allowed togo through up to one bar of a vertex) may be possible while keeping the areasub-quadratic. 14traight-line drawings of outer-1-planar graphs are also of interest. It isknown that there are order-preserving outer-1-planar straight-line drawings ofarea O ( n ) [2]. Are there straight-line drawings of sub-quadratic area (againperhaps at the expense of not respecting the 1-planar embedding)? References [1] E. Argyriou, S. Cornelsen, H. F¨orster, M. Kaufmann, M. N¨ollenburg,Y. Okamoto, C. Raftopoulou, and A. Wolff. Orthogonal and smooth or-thogonal layouts of 1-planar graphs with low edge complexity. In T. Biedland A. Kerren, editors,
Graph Drawing and Network Visualization (GD2018) , volume 11282 of
LNCS , pages 509–523. Springer, 2018.[2] C. Auer, C. Bachmaier, F. Brandenburg, A. Gleißner, K. Hanauer,D. Neuwirth, and J. Reislhuber. Outer 1-planar graphs.
Algorithmica ,74(4):1293–1320, 2016.[3] T. Biedl. Drawing outer-planar graphs in O ( n log n ) area. In S. Kobourovand M. Goodrich, editors, Graph Drawing (GD’01) , volume 2528 of
LNCS ,pages 54–65. Springer, 2002.[4] T. Biedl. Small drawings of outerplanar graphs, series-parallel graphs, andother planar graphs.
Discrete and Computational Geometry , 45(1):141–160,2011.[5] T. Biedl. Height-preserving transformations of planar graph drawings. InC. Duncan and A. Symvonis, editors,
Graph Drawing (GD’14) , volume8871 of
LNCS , pages 380–391. Springer, 2014.[6] T. Biedl. Drawing outer-1-planar graphs revisied. In
Graph Drawing andNetwork Visualization (GD’20) , LNCS. Springer, 2020. Poster with a shortabstract. To appear. See also ArXiV 2009.07106.[7] H. Dehkordi and P. Eades. Every outer-1-plane graph has a right anglecrossing drawing.
Int. J. Comput. Geom. Appl. , 22(6):543–558, 2012.[8] R. Diestel.
Graph Theory, 4th Edition , volume 173 of
Graduate texts inmathematics . Springer, 2012.[9] R. Eggleton. Rectilinear drawings of graphs.
Utilitas Mathematica , 29:149–172, 1986.[10] H. de Fraysseix, J. Pach, and R. Pollack. Small sets supporting Fary em-beddings of planar graphs. In
ACM Symposium on Theory of Computing(STOC ’88) , pages 426–433, 1988.[11] S. Hong, P. Eades, G. Liotta, and S. Poon. F´ary’s theorem for 1-planargraphs. In J. Gudmundsson, J. Mestre, and T. Viglas, editors,
Computingand Combinatorics (COCOON 2012) , volume 7434 of
LNCS , pages 335–346. Springer, 2012. 1512] S. Kobourov, G. Liotta, and F. Montecchiani. An annotated bibliographyon 1-planarity.