On the Perspectives Opened by Right Angle Crossing Drawings

Transcription

1 Journal of Graph Algorthms and Applcatons vol. 5, no., pp (20) On the Perspectves Opened by Rght Angle Crossng Drawngs Patrzo Angeln Luca Cttadn Guseppe D Battsta Walter Ddmo 2 Fabrzo Frat Mchael Kaufmann 3 Antonos Symvons 4 Dpartmento d Informatca e Automazone, Roma Tre Unversty, Italy 2 Dpartmento d Ingegnera Elettronca e dell Informazone, Peruga Unversty, Italy 3 Wlhelm-Schckard-Insttut für Informatk, Unverstät Tübngen, Germany 4 Department of Mathematcs, Natonal Techncal Unversty of Athens, Greece Submtted: December 2009 Artcle type: Regular paper Revewed: October 200 Fnal: November 200 Revsed: October 200 Publshed: February 20 Accepted: November 200 Communcated by: D. Eppsten and E. R. Gansner Work partally supported by the Italan Mnstry of Research, Grant number RBIP06BZW8, project FIRB Advanced trackng system n ntermodal freght transportaton. E-mal addresses: angeln@da.unroma3.t(patrzo Angeln) ratm@da.unroma3.t(luca Cttadn) gdb@da.unroma3.t (Guseppe D Battsta) walter.ddmo@de.unpg.t (Walter Ddmo) frat@da.unroma3.t (Fabrzo Frat) mk@nformatk.un-tuebngen.de (Mchael Kaufmann) symvons@math.ntua.gr (Antonos Symvons)

2 54 Angeln et al. RAC Drawngs Abstract Rght Angle Crossng (RAC) drawngs are polylne drawngs where each crossng forms four rght angles. RAC drawngs have been ntroduced because cogntve experments provded evdence that ncreasng the number of crossngs does not decrease the readablty of a drawng f edges cross at rght angles. We nvestgate to what extent RAC drawngs can help n overcomng the lmtatons of wdely adopted planar graph drawng conventons, provdng both postve and negatve results. Frst, we prove that there exst acyclc planar dgraphs not admttng any straght-lne upward RAC drawng and that the correspondng decson problem s NP-hard. Also, we show dgraphs whose straght-lne upward RAC drawngs requre exponental area. Explotng the technques ntroduced for studyng straght-lne upward RAC drawngs, we also show that there exst planar undrected graphs requrng quadratc area n any straght-lne RAC drawng. Second, we study whether RAC drawngs allow us to draw boundeddegree graphs wth lower curve complexty than the one requred by more constraned drawng conventons. We prove that every graph wth vertexdegree at most sx (at most three) admts a RAC drawng wth curve complexty two (resp. one) and wth quadratc area. Thrd, we consder a natural non-planar generalzaton of planar embedded graphs. Here we gve bounds for curve complexty and area dfferent from the ones known for planar embeddngs. Introducton In graph drawng, t s commonly accepted that crossngs and bends can make the layout dffcult to read and expermental results show that the human performance n path-tracng tasks s negatvely correlated to the number of edge crossngs and to the number of bends along the edges [20, 2, 23]. However, further cogntve experments n graph vsualzaton show that ncreasng the number of crossngs does not decrease the readablty of the drawng f the edges cross at rght angles [4, 5]. These results provde evdence for the effectveness of orthogonal drawngs (n whch edges are chans of horzontal and vertcal segments) wth few bends [5, 6] and motvate the study of a new class of drawngs, called Rght Angle Crossng drawngs (RAC drawngs), ntroduced by Ddmo, Eades, and Lotta [9]. A RAC drawng of a graph G s a polylne drawng Γ of G such that any two crossng segments are orthogonal. Fgure shows a RAC drawng wth curve complexty two, where the curve complexty of Γ s the maxmum number of bends along an edge of Γ. If Γ has curve complexty zero, then Γ s a straght-lne RAC drawng. Ths paper nvestgates RAC drawngs wth low curve complexty for both drected and undrected graphs. For drected graphs, also called dgraphs, a wdely studed drawng standard s the upward drawng conventon, where edges are monotone n the vertcal drecton. A dgraph has an upward planar drawng f and only f t has a straght-lne upward planar drawng [6]. However, not all planar dgraphs have

3 JGAA, 5() (20) 55 Fgure : A RAC drawng wth curve complexty two. an upward planar drawng and straght-lne upward planar drawngs requre exponental area for some famles of dgraphs [7]. We nvestgate straght-lne upward RAC drawngs,.e. straght-lne upward drawngs wth rght angle crossngs. In partcular, t s natural to ask f every planar acyclc dgraph admts an upward RAC drawng and f every dgraph wth an upward RAC drawng admts one wth polynomal area. Both these questons have a negatve answer: We prove that there exst acyclc planar dgraphs that do not admt any straght-lne upward RAC drawng and that the problem of decdng whether an acyclc planar dgraph admts such a drawng s NP-hard; we show that there exst upward planar dgraphs whose straght-lne upward RAC drawngs requre exponental area. Explotng the technques ntroduced for provng that straght-lne upward RAC drawngs of upward planar dgraphs may requre exponental area, we also show that there exst planar undrected graphs requrng quadratc area n any straght-lne RAC drawng. It s known [9] that any n-vertex straght-lne RAC drawng of an undrected graph has at most 4n 0 edges, for every n 4, and ths bound s tght. Further, every graph admts a RAC drawng wth at most three bends per edge, and ths curve complexty s requred n nfntely many cases [9]. Indeed, RAC drawngs wth curve complexty one and two have at most 2n and 50n edges, respectvely, as shown by Arkush and Tóth [], who mproved prevous sub-quadratc area bounds by Ddmo et al. [9]. Hence, we nvestgate famles of graphs that can be drawn wth curve complexty one or two, provng the followng results: Every degree-6 graph admts a RAC drawng wth curve complexty two; every degree-3 graph admts a RAC drawng wth curve complexty one. In both cases, the drawngs can be computed n lnear tme and requre quadratc area. Observe that degree-4 graphs, wth the excepton of the octahedron [2], admt planar orthogonal drawngs wth curve complexty two [8],

4 56 Angeln et al. RAC Drawngs whle there exst degree-3 graphs, as for example K 4, that requre two bends on one edge n any planar orthogonal drawng. In a fxed embeddng settng, the nput graph G s gven wth a (non-planar) embeddng,.e., a crcular orderng of the edges ncdent to each vertex and an orderng of the crossngs along each edge. A RAC drawng algorthm can not change the embeddng of G. For such a settng t has been proved [9] that any n-vertex graph admts a RAC drawng wth O(kn 2 ) bends per edge, where k s the maxmum number of crossngs between any two edges. Also, there exst graphs whose RAC drawngs requre Ω(n 2 ) bends along some edges. We study the fxed embeddng settng, namely we study non-planar graphs obtaned by augmentng a plane trangulaton wth edges nsde pars of adjacent faces; we call these graphs kte-trangulatons: We prove that one bend per edge s always suffcent and sometmes necessary for a RAC drawng of a kte-trangulaton; we show that there exst kte-trangulatons requrng cubc area n any straght-lne RAC drawng. Recall that every embedded planar graph admts a planar drawng wth quadratc area [4, 22]. The rest of the paper s organzed as follows. In Sect. 2 we ntroduce some defntons and prelmnares; n Sect. 3 we study straght-lne upward RAC drawngs of planar acyclc dgraphs; n Sect. 4 we study RAC drawngs of bounded-degree graphs; n Sect. 5 we study RAC drawngs of kte-trangulatons; fnally, n Sect. 6, we conclude the paper wth some open problems. 2 Prelmnares We assume famlarty wth graph drawng and planarty[5, 6]. In the followng, unless otherwse specfed, all consdered graphs are smple. The degree of a vertex s the number of edges ncdent to t. The degree of a graph s the maxmum among the degrees of ts vertces. A graph s regular f all ts vertces have the same degree. A drawng of a graph s a mappng of each vertex to a dstnct pont of the plane and of each edge to a Jordan curve between ts endponts. A straght-lne drawng s such that all edges are straght-lne segments. A polylne drawng s such that all edges are sequences of straght-lne segments, where any pont shared by consecutve segments of dfferent slopes s a bend. The curve complexty of a drawng Γ s the maxmum number of bends along an edge n Γ. A grd drawng of a graph s such that each vertex has nteger coordnates. The area of a grd drawng s the area of the smallest rectangle wth sdes parallel to the axes completely enclosng the drawng. A planar drawng s such that no two edges ntersect except, possbly, at common endponts. A planar drawng of a graph determnes a crcular orderng of the edges ncdent to each vertex. Two drawngs of the same graph are equvalent f they determne the same crcular orderng around each vertex. A planar embeddng s an equvalence class

5 JGAA, 5() (20) 57 of planar drawngs. A planar drawng parttons the plane nto topologcally connected regons, called faces. The unbounded face s the external face. A graph together wth a planar embeddng and a choce for ts external face s called plane graph. A plane graph s a trangulaton when all ts faces are trangles. When dealng wth non-planar graphs, an embeddng of such a graph s a crcular orderng of the edges ncdent to each vertex and a lnear order of the edges crossng each edge. An upward drawng of a dgraph s such that all edges are curves monotoncally ncreasng n the upward drecton. An upward planar drawng of a dgraph G s a drawng of G that s both upward and planar. If G admts an upward planar drawng, then G s an upward planar dgraph. A Rght Angle Crossng drawng (RAC drawng) of a graph G s a polylne drawng Γ of G such that any two crossng segments n Γ are orthogonal. If a RAC drawng Γ has curve complexty zero, then Γ s a straght-lne RAC drawng. An upward RAC drawng of a dgraph s a RAC drawng that s also upward. A fan n a drawng Γ s a par of edge segments ncdent to the same vertex. Two segments s and s 2 crossng the same segment n Γ are parallel. Ths leads to the followng propertes, llustrated n Fg. 2(a) and 2(b), and proved n [9] and [0]. v a u b c z (a) (b) Fgure 2: Illustratons for (a) Property and for (b) Property 2. Property In a straght-lne RAC drawng no edge can cross a fan. Property 2 In a straght-lne RAC drawng there can not be a trangle and two edges (a,b), (a,c) such that a les outsde and b,c le nsde. 3 Upward RAC Drawngs We now study straght-lne upward RAC drawngs of drected graphs. In order to acheve our results on straght-lne upward RAC drawngs of drected graphs, weprovesomelemmata concernngundrected graphs. ConsderK 4, that s, the complete graph on four vertces u, v, z, and w. Let E and E 2 be the embeddngs of K 4 shown n Fg. 3(a) and 3(b), respectvely. Lemma In any straght-lne drawng of K 4, ts embeddng s one of E and E 2, up to a renamng of the vertces.

6 58 Angeln et al. RAC Drawngs v v z w z w (a) u u (b) Fgure 3: (a) E ; (b) E 2. Proof: Consderanystraght-lnedrawngΓofK 4. Ether three orfourvertces are on the convex hull of Γ, as otherwse there would be two overlappng edges. Observe that, snce the drawng s straght-lne, the edges delmtng the convex hull of Γ do not cross any edge of K 4. If exactly three vertces of K 4 are on the convex hull of Γ, then the fourth vertex s nsde such a convex hull. Snce the drawng s straght-lne, the edges ncdent to the fourth vertex do not cross any edge of K 4. It follows that the embeddng of K 4 s E. If exactly four vertces of K 4 are on the convex hull of Γ, then the two edges between non-consecutve vertces of the convex hull cross. Snce the drawng s straght-lne, such edges cross exactly once. It follows that the embeddng of K 4 s E 2. Lemma 2 Let G be a graph contanng two vertex-dsjont copes K 4 and K 4 of K 4. Let Γ be any straght-lne RAC drawng of G. For any 3-cycle (a,b,c ) of K 4, whch s represented n Γ by a trangle, ether all the vertces of K 4 are nsde or they are all outsde t. Proof: If at least two vertces a and b of K 4 are nsde and at least one vertex c s outsde t, then Property 2 s volated, snce vertex c s connected to both a and b. b b a a b c c d a a b c c d d (a) d (b) Fgure 4: (a) If d s placed outsde, then Property 2 s volated. (b) If d s placed nsde, then Property 2 s volated.

7 JGAA, 5() (20) 59 If exactly one vertex a of K 4 s nsde, then b, c, and d are outsde t. Snce the drawng s straght-lne, f there s a crossng between an edge of the 3-cycle (b,c,d ) of K 4 and an edge of (a,b,c ), then such an edge of (b,c,d ) crosses a fan composed of two edges of (a,b,c ), thus volatng Property. It follows that s contaned nsde the trangle representng (b,c,d ) n Γ, wth each of the edges (a,b ), (a,c ), and (a,d ) crossng a dstnct edge of wth a rght-angle crossng. However, n ths case every possble placement of d volates Property 2. Namely, f d s outsde, then Property 2 s volated snce d s connected to a, b, and c, whch are nsde (see Fg. 4(a)). Further, f d s nsde, then t s nsde one of the faces nternal to, say the one contanng a ; then, Property 2 s volated snce d and a are both connected to b, whch s outsde the trangle representng such a face n Γ (see Fg. 4(b)). Lemma 3 Let G be a graph contanng two vertex-dsjont copes K 4 of K 4. In any RAC drawng Γ of G, no edge of K 4 crosses an edge of K 4. and K 4 Proof: Let Γ be Γ restrcted to the edges of K 4 and K 4. We show that n Γ there s no crossng between the edges of K 4 and the edges of K 4. Let u, v, z, and w be the vertces of K 4, and let u, v, z, and w be the vertces of K 4. If the embeddng of K 4 n Γ s E, then assume, wthout loss of generalty up to a renamng of the vertces, that (u,v,z ) s the 3-cycle delmtng the external face of K 4 n Γ and hence enclosng w. By Lemma 2, ether all the vertces of K 4 le outsde or they all le nsde t. In the former case, f there s a crossng between an edge of K 4 and an edge of K 4, then such an edge of K 4 crosses a fan composed of two edges of K 4, thus volatng Property. In the latter case, the vertces of K 4 le n the faces of K 4 nternal to. By Lemma 2, all the vertces of K 4 le n the same nternal face of K 4. Hence, n both cases, no edge of K 4 crosses an edge of K 4. If the embeddng of K 4 n Γ s E 2, then assume, wthout loss of generalty up to a renamng of the vertces, that (u,v,z,w ) s the 4-cycle delmtng the external face of K 4 n Γ. Thus, the edges of K 4 delmt fve connected regons R,...,R 5 of the plane, where R,R 2,R 3, and R 4 are nsde (u,v,z,w ), and R 5 s outsde (u,v,z,w ). We prove that all the vertces of K 4 are nsde the same regon R. Suppose that vertces a and b exst such that a,b {u,z,v,w } and a s nsde R and b s nsde R j, wth j. For every par of regons R and R j, wth j, a 3-cycle (a,b,c ) of K 4, wth a,b,c {u,z,v,w }, exsts contanng R n ts nteror and R j n ts exteror, or vce versa. Then, a s nsde the trangle representng (a,b,c ) and b s outsde such a trangle, or vce versa. However, by Lemma 2, Γ s not a RAC drawng. Hence, all the vertces of K 4 are nsde the same regon R. If all the vertces of K 4 are n the same regon R, wth 4, then no edge of K 4 crosses an edge of K 4. If all the vertces of K 4 are n R 5, then suppose that a crossng between an edge of K 4 and an edge of K 4 exsts. However, such an edge of K 4 crosses a fan composed of two edges of K 4, thus volatng Property.

8 60 Angeln et al. RAC Drawngs v z w Fgure 5: The upward planar dgraph H obtaned by acyclcally orentng the edges of K 4. u Now we use the prevous lemmata to prove the man results of ths secton. Frst, we ntroduce an upward planar dgraph H, shown n Fg. 5, whch s obtaned by acyclcally orentng the edges of K 4. Denote by u and v the only source and the only snk of H, respectvely. We get the followng: Lemma 4 Consder a planar acyclc dgraph K. Replace each edge (a,b) of K wth a copy of H, by dentfyng vertces a and b of K wth vertces u and v of H, respectvely. Let K be the resultng planar dgraph. Dgraph K s upward planar f and only f K s straght-lne upward RAC drawable. Proof: Refer to Fgs. 6(a) and 6(b). Frst, suppose that K admts an upward planar drawng. Then, by the results of D Battsta and Tamassa [6], K admts a straght-lne upward planar drawng Γ. Consder the drawng Γ of K obtaned from Γ by drawng each copy of H that replaces an edge (a,b) n such a way that: () The drawng of H s upward planar; () the drawng of edge (u,v) of H n Γ concdes wth the drawng of edge (a,b) of K n Γ; and () the drawng of the other vertces and edges of H s arbtrarly close to (u,v). Snce Γ s a straght-lne upward planar drawng, Γ s a straght-lne upward planar drawng. Hence, Γ s a straght-lne upward RAC drawng of K. Second, suppose that K admts a straght-lne upward RAC drawng Γ. Consder the straght-lne drawng Γ of K obtaned by restrctng Γ to the edges of K, that s, obtaned by removng from Γ, for every copy of H, all the vertces of H dfferent from u and v and all the edges of H dfferent from (u,v). As Γ s an upward drawng of K, then Γ s an upward drawng of K. Suppose, for a contradcton, that two edges cross n Γ. If such two edges are adjacent, then they do not cross, as otherwse they overlap. If such two edges are not adjacent, then they belong to two dstnct copes of H n K. However, by Lemma 3, no two edges belongng to dstnct copes of H cross n Γ, thus obtanng a contradcton. Hence, Γ s a straght-lne upward planar drawng of K. We are ready to prove the frst theorem of ths secton.

9 JGAA, 5() (20) (a) (b) Fgure 6: (a) A straght-lne upward drawng of an upward planar acyclc dgraph K. (b) A straght-lne upward RAC drawng of the planar acyclc dgraph K obtaned by replacng each edge of K wth a copy of H (a) (b) Fgure 7: (a) A planar acyclc dgraph G that s not upward planar. (b) The planar acyclc dgraph G obtaned by replacng each edge of G wth a copy of H s not straght-lne upward RAC drawable. Theorem There exst acyclc planar dgraphs that do not admt any straghtlne upward RAC drawng. Proof: Consder any planar acyclc dgraph G (as the one of Fg. 7(a)) that s not upward planar. By Lemma 4, the planar acyclc dgraph G obtaned by replacng each edge of G wth a copy of H does not admt any straght-lne upward RAC drawng (see Fg. 7(b)). Note that there exst planar dgraphs, as the one n Fg. 8, that do not admt any straght-lne upward RAC drawng, that are not constructed usng gadget H, and whose sze s smaller than the one of the dgraph n Fg. 7(b). However, provng that they are not straght-lne upward RAC drawable could result n a complex case-analyss. Motvated by the fact that there exst acyclc planar dgraphs that do not admt any straght-lne upward RAC drawng, we study the tme complexty of the correspondng decson problem. We show that the problem of testng whether a dgraph admts a straght-lne upward RAC drawng (Upward RAC Drawablty Testng) s NP-hard, by

10 62 Angeln et al. RAC Drawngs Fgure 8: An 8-vertex planar dgraph that does not admt any straght-lne upward RAC drawng. means of a reducton from the problem of testng whether a dgraph admts a straght-lne upward planar drawng (Upward Planarty Testng), whch s NP-complete [3]. Theorem 2 Upward RAC Drawablty Testng s NP-hard. Proof: We reduce Upward Planarty Testng to Upward RAC Drawablty Testng. Let G be an nstance of Upward Planarty Testng. Replace each edge (a,b) of G wth a copy of H, by dentfyng vertces a and b of G wth vertces u and v of H, respectvely. Let G be the resultng planar dgraph. By Lemma 4, G s upward planar f and only f G admts a straght-lne upward RAC drawng. Next, we show that there exsts a class of planar acyclc dgraphs that requre exponental area n any straght-lne upward RAC drawng. Consder the class of upward planar dgraphs G n (see Fg.9), defned by D Battsta et al. [7], whch requres Ω(2 n ) area n any straght-lne upward planar drawng,underanyresolutonrule. Replaceeachedge(a,b) ofg n wth acopyof H, by dentfyng vertces a and b of G n wth vertces u and v of H, respectvely. Let G n be the resultng planar dgraph. Observe that, assumng that G n has n vertces, G n has O(n) vertces snce, for every edge of G n, two new vertces are ntroduced n G n. Theorem 3 Any straght-lne upward RAC drawng of G n requres Ω(b n ) area, under any resoluton rule, for some constant b >. Proof: Suppose, for a contradcton, that, for every constant b >, G n admts a straght-lne upward RAC drawng Γ wth o(b n ) area, under some resoluton rule. Consder the straght-lne drawng Γ of G n obtaned by restrctng Γ to the edges of G n, that s, obtaned by removng from Γ, for every copy of H, all the vertces of H dfferent from u and v and all the edges of H dfferent from (u,v). As Γ s an upward drawng of G n, then Γ s an upward drawng of G n. If two edges of G n are adjacent, then they do not cross n Γ, as otherwse they overlap. If two edges of G n are not adjacent, then they belong to two dstnct

11 JGAA, 5() (20) 63 t n t n t G n G n t 0 t 0 G G 0 s 0 s n s 0 s s n (a) (b) (c) Fgure 9: (a) Graph G 0. (b) Graph G. (c) Graph G n. copes of H n G n. However, by Lemma 3, no two edges belongng to dstnct copes of H cross n Γ, thus they do not cross n Γ. Hence, Γ s a straght-lne upward planar drawng of G n. Further, the area of Γ s o(b n ), as the area of Γ s o(b n ), thus obtanng a contradcton and provng the theorem. We now turn our attenton to straght-lne RAC drawngs of undrected graphs. We explot the technques ntroduced for straght-lne upward RAC drawngs to get a quadratc lower bound on the area requrements of straghtlne grd RAC drawngs of planar graphs. Consder a nested trangles graph G, that s, a trconnected graph composed of n 3 3-cycles nested one nto the other (see Fg. 0(a)). Graph G s known to requre Ω(n 2 ) area n any straght-lne planar drawng [4]. Replace each edge (a,b) of G wth a copy of K 4, by dentfyng vertces a and b of G wth vertces u and v of K 4, respectvely. Let G be the resultng planar graph (see Fg. 0(b)). Observe that G has O(n) vertces snce, for every edge of G, two new vertces are ntroduced n G. We have the followng. Theorem 4 Any straght-lne grd RAC drawng of G requres Ω(n 2 ) area. Proof: Consder any straght-lne grd RAC drawng Γ of G. Consder the straght-lne drawng Γ of G obtaned by restrctng Γ to the edges of G, that s, obtaned by removng from Γ, for every copy of K 4, all the vertces of K 4 dfferent from u and v and all the edges of K 4 dfferent from (u,v). If two edges of G are adjacent, then they do not cross, as otherwse they overlap. If two edges of G are not adjacent, then they belong to two dstnct copes of K 4 n G. However, by Lemma 3, no two edges belongng to dstnct copes of K 4 cross n Γ, thus they do not cross n Γ. Hence, Γ s a straght-lne planar drawng of G. It follows that the area of Γ s Ω(n 2 ), and the area of Γ s Ω(n 2 ), as well.

12 64 Angeln et al. RAC Drawngs (a) (b) Fgure 0: (a) A nested trangles graph G. (b) The graph G obtaned by replacng each edge (a,b) of G wth a copy of K 4. 4 RAC-Drawngs of Bounded-Degree Graphs In ths secton, we present algorthms for constructng RAC drawngs of graphs of bounded degree. The algorthms are based on the decomposton of a regular drected multgraph nto drected 2-factors. A 2-factor of an undrected graph G s a spannng subgraph of G consstng of vertex-dsjont cycles (see also [3, pp.227]). Analogously, a drected 2-factor of a drected graph s a spannng subgraph consstng of vertex-dsjont drected cycles. The decomposton of a regular drected multgraph nto drected 2-factors follows from a classcal result for undrected graphs [9] statng that a regular multgraph of degree 2k has k edge-dsjont 2-factors. A constructve proof of the followng theorem was gven by Eades et al. []. Theorem 5 (Eades,Symvons,Whtesdes []) Let G = (V, E) be an n- vertex undrected graph of degree and let d = /2. Then, there exsts a drected mult-graph G = (V,E ) such that:. each vertex of G has ndegree d and outdegree d; 2. G s a subgraph of the underlyng undrected graph of G ; and 3. the edges of G can be parttoned nto d edge-dsjont drected 2-factors. Furthermore, the drected graph G and ts d drected 2-factors can be computed n O( 2 n) tme. Let u be a vertex placed at a grd pont. We say that an edge e extng u uses the Y-port of u (resp. the Y-port of u) f t exts u along the +Y drecton (resp. along the Y drecton). In an analogous way, we defne the X-port and the X-port. We have the followng. Theorem 6 Every n-vertex graph wth degree at most sx admts a RAC drawng wth curve complexty two n O(n 2 ) area. Such a drawng can be computed n O(n) tme.

13 JGAA, 5() (20) 65 Proof: Let G = (V,E) be a graph of degree sx. Let G = (V,E ) be the drected multgraph obtaned from G as n Theorem 5, and let C,C 2, and C 3 be the three edge-dsjont drected 2-factors of G. We show how to obtan a RAC drawng of G. Then, a RAC drawng of G can be obtaned by removng from the drawng all the edges n E \ E and by gnorng the drecton of the edges (a) (b) Fgure : (a) A regular drected multgraph G wth ndegree and outdegree equal to three and ts drected 2-factors C, C 2, and C 3. The edges of C are represented by sold thn lnes, the edges of C 2 are represented by sold thck lnes, andthe edgesofc 3 arerepresentedbydashedlnes. (b) The RACdrawng of G wth two bends per edge constructed by the algorthm descrbed n the proof of Theorem 6. ThealgorthmplacesthevertcesofV onthemandagonalofano(n) O(n) grd, n an order determned by one of the drected 2-factors, say C. Most of the edges of C are drawn as straght-lne segments along the dagonal whle the edges of C 2 and C 3 are drawn as 3-segment lnes above and below the dagonal, respectvely. Fnally, the remanng closng edges of C (.e., the edges that are not drawn on the dagonal) are drawn as 2- or 3-segment lnes ether above or below the dagonal. We frst descrbe how to place the vertces of G along the man dagonal. Arbtrarly name the cycles c,c 2,...,c k of C. Consder each cycle c, for k. If there exst a vertex u c and an edge (u,z) C 2 or C 3 such that z belongs to a cycle c j of C wth j >, then let u be the topmost vertex of c and let the vertex followng u n c be the bottommost vertex of c. Otherwse, fthere exst a vertexv c and an edge(v,w) C 2 orc 3 such that w belongs to a cycle c j of C wth j <, then let v be the bottommost vertex of c and let the vertex precedng v n c be the topmost vertex of c. Otherwse, all the edges of C 2 and C 3 extng vertces of c are drected to vertces of c. In ths case, let an arbtrary vertex w of c be the

14 66 Angeln et al. RAC Drawngs bottommostvertexofc andletthevertexprecedngwnc bethetopmost vertex of c. Fgure (a) shows a regular drected multgraph G of ndegree and outdegree three and ts drected 2-factors C, C 2, and C 3. C conssts of cycles c : (5,,2,3,4,5) and c 2 : (6,7,8,9,6). We set 4 as the topmost vertex of c snce edge (4,6) of C 2 has vertex 6 of c 2 as ts destnaton. Analogously, we set 6 as the bottommost vertex of c 2 snce edge (6,5) of C 2 has vertex 5 of c as ts destnaton. Fgure (b) shows the RAC drawng of G wth curve complexty two constructed by the algorthm descrbed n ths proof. Then, the vertces of G are placed on the dagonal so that each vertex of c s placed on the dagonal before each vertex of c j, for each < j, and so that the vertces of c are placed on the dagonal n the order defned by c, startng at the bottommost vertex of c and endng at the topmost vertex of c, for each. When the h-th vertex of G s placed on the dagonal, t s assgned coordnates (6(h ),6(h )). Havng placed the vertces on the grd, we turn our attenton to drawng the edges of G. Each edge s drawn ether as a -segment lne along the dagonal, or as a 2- or 3-segment lne ether above or below the dagonal. We draw the edges so that all the crossng lne segments are parallel to the axes and, consequently, all the crossngs are at rght angles. In our drawngs, every lne segment s that s not parallel to the axes s ncdent to a vertex v s of the graph; further, such a segment s s contaned n a dedcated regon wthn a square Q(v s ) whose dagonals meet at v s and whose sde has length 6 (see Fg. 2(a)). The edges of C 2 are drawn above the dagonal as follows. Consder an edge (u,v) of C 2 and let u and v be placed at grd ponts (u x,u y ) and (v x,v y ), respectvely. If u s placed below v (.e., u y < v y ), then edge (u,v) s drawn as a 3- segment lne extng vertex u from the Y-port and beng defned by bendponts (u x,v y 4) and (v x 5,v y 4). Note that the thrd lne segment of (u, v) s contaned n the lghtly-shaded regon (above the dagonal) of the south-west quadrant of Q(v) (see Fg. 2(a)). If u s placed above v (.e., u y > v y ), then edge (u,v) s drawn as a 3-segment lne extng vertex u from the X-port and beng defned by bend-ponts (v x +3,u y ) and (v x +3,v y +4). Note that, n ths case, the thrd lne segment of(u, v) s contaned n the lghtly-shaded regon(above the dagonal) of the north-east quadrant of Q(v) (see Fg. 2(a)). It s easy to observe that the only lne segments that belong to edges of C 2 and that cross other lne segments are parallel to the axes, hence they cross at rght angles. Namely, all the lne segments that are not parallel to the axes are contaned n the lghtly-shaded regons shown n Fg. 2(a), and there s at most one of such lne segments per regon. The edges of C 3 are drawn below the dagonal n an analogous way.

15 JGAA, 5() (20) 67 v +8 y C C2 C 3 vy C v C u C 2 v y 8 v x 8 C 3 C vx (a) v +8 x v (b) Fgure 2: (a) The square Q(v) around a vertex v. The shaded regons contan lne segments not parallel to the axes and are used to vsualze the absence of crossngs nsde Q(v). (b) Drawng the closng edge of a cycle of C n Case 3. Consder now the edges of C. All such edges, except those closng the cycles of C, are drawn as straght-lne segments along the dagonal. As all the edges of C 2 (resp. C 3 ) are drawn above (resp. below) the dagonal, the edges of C drawn along the dagonal are not nvolved n any edge crossng. To complete the drawng of G, we descrbe how to draw the edges connectng the topmost vertex to the bottommost vertex of each cycle of C. Consder an arbtrary cycle c of C and let (u,v) be ts closng edge. We consder three cases: Case : u was selected to be the topmost vertex of c due to the exstence of an edge (u,z) of C 2 or C 3 such that z s above u. In such a case, after drawng the edges of C 2 and C 3, vertex u has not used ether ts X-port, or ts Y-port, or both. Namely, u used ts X-port f there s an edge (u,v) of C 2 such that v s below u, and u used ts Y-port f there s an edge (u,v) of C 3, such that v s below u. However, snce an edge (u,z) of C 2 or of C 3 exsts such that z s above u, f u used both ts X-port and ts Y-port, there would be three edges extng u n C 2 and C 3, whle there are exactly two of such edges. Assume that the X-port of u s free (the case where the Y-port of u s free can be treated analogously). Edge (u,v) s drawn above the dagonal as a 3-segment lne extng vertex u from the X-port and beng defned by bendponts (v x +,u y ) and (v x +,v y + 7). Note that, n ths case, the thrd lne segment of (u, v) s contaned n the dark-shaded regon (above the dagonal) of the north-east quadrant of Q(v) (see Fg. 2(a)). Case 2: v was selected to be the bottommost vertex of c due to the exstence of an edge (v,w) of C 2 or C 3 such that w s below v. In such a case, after drawng the edges of C 2 and C 3, vertex v has not used ether ts X-port, or ts Y-port, or both, whch can be proved analogously to

16 68 Angeln et al. RAC Drawngs Case. Assume that the Y-port of v s free (the case where the X-port of v s free can be treated analogously). Edge (u,v) s drawn above the dagonal as a 3- segment lne extng vertex v from the Y-port and beng defned by bend-ponts (v x,u y ) and (u x 7,u y ). Note that, n ths case, the frst lne segment of (u,v) s contaned n the dark-shaded regon (above the dagonal) of the south-west quadrant of Q(u) (see Fg. 2(a)). Case 3: Nether Case nor Case 2 apples. In such a case, all the edges of C 2 and C 3 extng vertces of cycle c are also drected to vertces of c. Notce that ths also mples that all the edges of C 2 and C 3 enterng vertces of c are orgnated from vertces of c. Namely, f there were an edge (u,v) such that v s n c and u s not, then there would be an edge (w,z) such that w s n c and z s not. Hence, denotng by u and v the topmost vertex and the bottommost vertex of c, respectvely, (observe that the bottommost vertex was chosen arbtrarly) the drawng of the edges of C 2 and C 3 ncdent to vertces of c takes place entrely wthn the square havng ponts(v x,v y )and(u x,u y )asoppostecorners(the shadedsquarenfg.2(b)). Hence, the closng edge can be drawn as a 2-segment lne connectng u and v and beng defned by bend-pont (v x,u y +) (see Fg. 2(b)). Gven C, C 2, and C 3, t s easy to see that the drawng can be constructed n lnear tme. By Theorem 5, C, C 2, and C 3 can be also computed n lnear tme, resultng n a lnear-tme algorthm. Also, the produced RAC drawng les n an O(n 2 ) sze grd. We now prove the followng: Theorem 7 Every n-vertex graph wth degree at most three admts a RAC drawng wth curve complexty one n O(n 2 ) area. Such a drawng can be computed n O(n) tme. Proof: Let G = (V,E) be a graph of degree three. Let G = (V,E ) be the drected multgraph obtaned from G as n Theorem 5. Observe that G s a regular multgraph of degree four. Let C and C 2 be two edge-dsjont drected 2-factors of G. We wll show how to obtan a RAC drawng of G such that only the edges of E and the edges of E \E mght partally overlap. Removng from the constructed drawng the edges of E \E results nto a RAC drawng of G. We place the vertces of G along the man dagonal of an O(n) O(n) grd based on ther orderof appearance alongthe cycles of C. Consderan arbtrary cycle c of C. If c contans an edge (u,v) E \E, then we make vertces u and v be the topmost and bottommost vertex of c, respectvely. Otherwse, f there exst a vertex u c and an edge (u,z) C 2 E such that z belongs to a cycle c j of C wth j >, then let u be the topmost vertex of c and let the vertex followng u n c be the bottommost vertex of c.

17 JGAA, 5() (20) (a) (b) (c) Fgure 3: (a) A graph G = (V,E) of degree three. (b) The regular drected multgraph G = (V,E ) wth ndegree and outdegree equal to two obtaned from G and ts drected 2-factors C and C 2. The edges of C are represented by sold lnes and the ones of C 2 by dashed lnes. Edges not n G are thnner than the other edges. (c) The RAC drawng of G wth one bend per edge constructed by the algorthm descrbed n the proof of Theorem 7. Otherwse, f there exst a vertex v c and an edge (v,w) C 2 E such that w belongs to a cycle c j of C wth j <, then let v be the bottommost vertex of c and let the vertex precedng v n c be the topmost vertex of c. Otherwse, all the edges of C 2 E extng vertces of c are also drected to vertces of c. In ths case, let an arbtrary vertex w of c be the bottommostvertexofc andletthevertexprecedngwnc bethetopmost vertex of c. Fgure 3(a) shows a graph G of degree three. Fgure 3(b) shows ts correspondng drected graph G and ts drected 2-factors C and C 2. C conssts of cycles c : (,2,3,) and c 2 : (4,5,6,4). We set 3 as the topmost vertex of c snce edge (3,5) of C 2 has vertex 5 of c 2 as ts destnaton. Analogously, we set 4 as the bottommost vertex of c 2 snce edge (4,2) of C 2 has vertex 2 of c as ts destnaton. Fgure 3(c) shows the RAC drawng of G wth curve complexty one constructed by the algorthm descrbed n ths proof. In such a drawng, edge overlaps are allowed nvolvng at least one edge n E \E. Then, the vertces of G are placed on the dagonal so that each vertex of c s placed on the dagonal before each vertex of c j, for each < j, and so that the vertces of c are placed on the dagonal n the order defned by c, startng at the bottommost vertex of c and endng at the topmost vertex of c, for each. When the h-th vertex of G s placed on the dagonal, t s assgned coordnates (2(h ),2(h )). Havng placed the vertces on the grd, we turn our attenton to drawng the edges of G. Each edge s drawn ether as a -segment lne along the dagonal, or as a 2-segment lne ether above or below the dagonal. We draw the edges so that all the crossng lne segments are parallel to the axes and, consequently, all the crossngs are at rght angles.

18 70 Angeln et al. RAC Drawngs We frst descrbe how to draw the edges of C 2. Consder an arbtrary edge (u,v) of C 2. If u s placed below v (.e., u y < v y ), then edge (u,v) s drawn as a 2- segment lne below the dagonal, extng vertex u from the X-port and beng defned by bend-pont (v x,u y ). Such a lne enters v from ts Yport. If u s placed above v (.e., u y > v y ), then edge (u,v) s drawn as a 2- segment lne above the dagonal, extng vertex u from the X-port and beng defned by bend-pont (v x,u y ). Such a lne enters vertex v from ts Y-port. The edges of C 2 do not overlap each other. Further, they ntersect each other only at rght angles, as every lne segment s parallel to the axes. Consder now the edges of C. All such edges, except those closng the cycles of C, are drawn as straght-lne segments along the dagonal. As each edge of C 2 s drawn above or below the dagonal, the edges of C drawn along the dagonal are not nvolved n any edge crossng. To complete the drawng of G, we descrbe how to draw the closng edge of each cycle of C. Consder an arbtrary cycle c of C and let (u,v) be ts closng edge. We consder four cases: Case : Edge (u,v) belongs to E \E. In ths case, (u,v) s not part of G and t s not n the drawng. Case 2: Edge (u,v) belongs to E and u was selected to be the topmost vertex of c due to the exstence of an edge (u,z) C 2 E such that z s above u. Snce both edges of c ncdent to u and edge (u,z) belong to G and snce there are at most three edges ncdent to u n G, both the X-port and the Y-port of u are free. Now observe that, snce both edges of c ncdent to v belong to G, then at most one of the two edges of C 2 ncdent to v belongs to G. Hence, at most one of the Y-port and the X-port of v s used by an edge of G (the other port mght be used by an edge that belongs to G but not to G). Thus, t s always possble to draw edge (u,z) wth ts only bend ether at pont (v x,u y ) or at pont (u x,v y ), so that t overlaps only wth an edge of E \E. Case 3: Edge (u,v) belongs to E and v was selected to be the bottommost vertex of c due to the exstence of an edge (v,w) E C 2 wth vertex w beng placed lower on the dagonal than v. Analogously to the prevous case, both the X-port and the Y-port of v are free and at most one of the two edges of C 2 ncdent to u belongs to G. Hence, at most one of the Y-port and the X-port of u s used by an edge of G and t s always possble to draw edge (v,w) wth ts only bend ether at pont (v x,u y ) or at pont (u x,v y ), so that t overlaps only wth an edge of E \E. Case 4: None of the above cases apples. In ths case, all the edges of C 2 E extng vertces of c are also drected to vertces of c. Notce that ths also mples that all the edges of C 2 E enterng vertces of c are orgnated from vertces of c. Hence, denotng by u and v the topmost vertex and the bottommost vertex of c, respectvely, (observe that the

19 JGAA, 5() (20) 7 bottommost vertex was chosen arbtrarly) the drawng of the edges of C 2 E ncdent to vertces of c takes place entrely wthn the square havng ponts (v x,v y ) and (u x,u y ) as opposte corners. Hence, the closng edge (u,v) of c can be drawn as a 2-segment lne connectng u and v and beng defned by bend-pont (v x,u y +). Gven C and C 2, t s easy to see that the drawng can be constructed n lnear tme. By Theorem 5, C and C 2 can be also computed n lnear tme, resultng n a lnear-tme algorthm. Also, the produced RAC drawng les n an O(n 2 ) sze grd. 5 RAC Drawngs of Kte-Trangulatons In ths secton we study the mpact of admttng orthogonal crossngs on the drawablty of the non-planar graphs obtaned by addng edges to maxmal planar graphs nsde two adjacent faces, n a fxed embeddng scenaro. We show that such graphs always admt RAC drawngs wth curve complexty one and that such a curve complexty s sometmes requred. Let G be a trangulaton and let (u,z,w) and (v,z,w) be two adjacent faces of G sharng edge (z,w). We say that [u,v] s a par of opposte vertces wth respect to (z,w). Let E + = {[u,v ] =,2,,k} be a set of pars of opposte vertces of G, where [u,v ] s a par of opposte vertces wth respect to (z,w ) and edge (u,v ) does not belong to G. Suppose that, for any,j k and j, edges (z,w ) and (z j,w j ) are not ncdent to the same face of G. Let G be the embedded non-planar graph obtaned by addng an edge (u,v ) to G, for each par [u,v ] n E +, so that edge (u,v ) crosses edge (z,w ) and does not cross any other edge of G. We say that G s a kte-trangulaton and that G s ts underlyng trangulaton. Fgure 4 shows a kte-trangulaton. We get the followng: Fgure 4: A kte-trangulaton G. Sold lnes represent the edges of the underlyng trangulaton G of G. Dashed lnes represent edges between pars of opposte vertces. Theorem 8 Every kte-trangulaton admts a RAC drawng wth curve complexty one.

20 72 Angeln et al. RAC Drawngs Proof: Consder any kte-trangulaton G and ts underlyng trangulaton G. Remove from G all the edges (z,w ), for =,...,k, obtanng a new planar graph G. Snce, by defnton, no two edges (z,w ) and (z j,w j ), wth,j k and j, are adjacent to the same face of G, all the faces of G contan at most four vertces. Construct any straght-lne planar drawng Γ of G. We show how to nsert n Γ edges (u,v ) and (z,w ), for each =,...,k, n order to obtan a RAC drawngγofg. We consdertwocases, dependngon whetherface(u,w,v,z ) s strctly convex n Γ or not. S( w ) S( z, w ) u S( z ) w p z (a) v l( w ) l( z ) S( u ) u p l( u ) w l( w ) l( v ) l( z z ) S( u v, ) v S( v ) (b) Fgure 5: Drawng (u,v ) and (z,w ) nsde (u,w,v,z ), f (u,w,v,z ) s strctly convex. (a) u les nsde S(z,w ); (b) both u and v le outsde S(z,w ). Suppose that (u,w,v,z ) s strctly convex n Γ. Consder the straghtlne segment z w and consder the lnes l(z ) and l(w ) orthogonal to z w and passng through z and through w, respectvely. Further, consder the followng three regons of the plane: The closed half-plane S(z ) delmted by l(z ) and not contanng w, the closed half-plane S(w ) delmted by l(w ) and not contanng z, and the open strp S(z,w ) delmted by l(z ) and l(w ). If at least one out of u and v, say u, les nsde S(z,w ) (see Fg. 5(a)), then draw edge (z,w ) as a straght-lne segment z w. Draw a straght-lne segment u p startng at u, orthogonally crossng(z,w ), and endng at a pont p arbtrarly close to (z,w ). Complete a drawng of (u,v ) by drawng the straght-lnesegmentp v. Ifboth u and v le outsde S(z,w ) (see Fg. 5(b)), by the strct convexty of (u,w,v,z ), u and v le one n S(z ) and one n S(w ) and segment u v ntersects segment z w. Hence, the open strp S(u,v ) delmted by the lnes l(u ) and l(v ) orthogonal to u v and passng through u and through v, respectvely, contans both w and z. Then, draw edge (u,v ) as a straght-lne segment u v ; draw a straght-lne segment w p startng at w, orthogonally crossng (u,v ), and endng at a pont p arbtrarly close to (u,v ); fnally, complete a drawng of (w,z ) by drawng the straght-lne segment p z. Supposethat(u,w,v,z )snotstrctlyconvex(seefg.6); moreprecsely,

21 JGAA, 5() (20) 73 p w p z u v Fgure 6: Drawng (u,v ) and (z,w ) nsde (u,w,v,z ), f (u,w,v,z ) s strctly convex. suppose that angle û z v 80, the cases n whch the angle greater than or equal to 80 s ncdent to another vertex beng analogous. Segment z w splts (u,w,v,z ) nto two trangles (u,z,w ) and (v,z,w ). Snce û z v 80, û z w 90 or ŵ z v 90. Suppose that û z w 90, the other case beng analogous. Consder a pont p nsde (u,w,v,z ), arbtrarly close to w. Draw edge (u,v ) as a polygonal lne composed of segments u p and p v. Snce û z w 90, the lne through z orthogonally crossng the lne through u and w crosses segment u w n an nteror pont. Hence, f p s suffcently close to w, a straght-lne segment z p can be drawn startng at z, orthogonally crossng segment u p, and endng at a pont p arbtrarly close to u p. Complete a drawng of (w,z ) by drawng the straght-lne segment p w. Theorem 9 There exst kte-trangulatons that do not admt any straght-lne RAC drawng. v a u x y z Fgure 7: An embedded graph that s a subgraph of nfntely many ktetrangulatons wth curve complexty one n any RAC drawng. Proof: Consder an embedded planar graph H defned as follows. Graph H has externalface (u,v,z). Let a be a vertex ofh creatngfaces (u,v,a) and (a,v,z). Let x and y be vertces of H creatng faces (u,a,x) and (y,a,z), respectvely, n such a way that [v,x] s a par of opposte vertces wth respect to edge (a,u) and that [v,y] s a par of opposte vertces wth respect to edge (a,z). See Fg. 7.

22 74 Angeln et al. RAC Drawngs Consder any kte trangulaton G contanng H as a subgraph and contanng edges (v,x) and (v,y), respectvelycrossng(a,u) and (a,z). Consder any RAC drawng Γ of G. Consder the trangle representng (a,u,z). Vertex v, whch les outsde, s connected to vertces x and y, whch le nsde. Hence, by Property 2, Γ can not be a straght-lne RAC-drawng of G. Planar graphs are a proper subset of straght-lne RAC drawable graphs. However, whle straght-lne planar drawngs can always be realzed on a grd of quadratc sze (see, e.g., [4, 22]), straght-lne RAC drawngs may requre larger area, as shown n the followng. Theorem 0 There exsts an n-vertex kte-trangulaton that requres Ω(n 3 ) area n any straght-lne grd RAC drawng. un un u n 6 un u 3 u n 2 un 5 u 2 u u n 2 u 2 u un 4 u n 3 u n 4 un 5 (a) un 6 un (b) u n 3 Fgure 8: (a) A kte trangulaton G requrng Ω(n 3 ) area n any straght-lne grd RAC drawng. (b) A straght-lne RAC drawng of G. Proof: Consder a trangulaton G defned as follows (see Fg. 8(a)). Let C = (u,u 2,...,u n 4,u n 3 ) be a smple cycle, for some odd nteger n. Insert a vertex u n 2 nsde C and connect t to u, wth =,2,...,n 3. Insert two vertces u n and u n outsde C. Connect u n to u, wth =,2,...,n 6, and connect u n to u n 3 ; connect u n to u n 6, u n 5, u n 4, u n 3, and u n. Let (u n 3,u n,u n ) be the external face of G. Let G be the kte-trangulaton obtaned from G by addng edges (u,u +2 ), for =,3,5,...,n 6, and edge (u,u n 4 ), so that (u,u +2 ) crosses edge (u +,u n 2 ) of G, and so that (u,u n 4 ) crosses edge (u n 3,u n 2 ) of G. In the followng we prove that, n any straght-lne RAC drawng of G, cycle C = (u,u 3,...,u n 6,u n 4,u ) s a strctly-convex polygon. Ths clam, together wth the observaton that G admts a straght-lne RAC drawng (see Fg. 8(b)), clearly mples the theorem, snce any strctly-convex polygon needs cubc area f ts vertces have to be placed on a grd (see, e.g., [2]). Suppose, for a contradcton, that there exsts a straght-lne RAC drawng Γ of G wth an angle u u +2 u nsde C. Then, any two segments

23 JGAA, 5() (20) 75 orthogonallycrossng u u +2 and u +2 u +4, respectvely, meet at a pont outsde C, possblyatnfnty, whletheyshouldmeetatu n 2,whchsnsdeC. Thus, ether u n 2 u + s not orthogonal to u u +2 or u n 2 u +3 s not orthogonal to u +2 u +4, hence contradctng the assumpton that Γ s a RAC drawng. 6 Conclusons and Open Problems When a graph G does not admt any planar drawng n some desred drawng conventon, requrng that all crossngs form rght angles can be consdered as an alternatve soluton for the readablty of a drawng of G. In ths drecton, ths paper has shown negatve results for drected graphs that must be drawn upward wth straght-lne edges, and postve results for undrected graphs that must be drawn wth edges bendng once or twce. We now lst some open problems that are related to the results of ths paper. Whle recognzng upward planar dgraphs s NP-hard, a characterzaton s known[6]statngthatadgraphsupwardplanarfandonlyftsasubgraphof a planar st-dgraph. As we proved that recognzng straght-lne upward RACdrawable dgraphs s also NP-hard, the followng problem naturally arses. Problem Is t possble to characterze dgraphs admttng straght-lne upward RAC drawngs? We have proved the exstence of nfntely many planar acyclc dgraphs not admttng any straght-lne upward RAC drawng. However, we are not aware of planar acyclc dgraphs requrng more than one bend on some edges. Problem 2 Does every planar acyclc dgraph admt an upward RAC drawng wth curve complexty one (wth curve complexty two)? There exst outerplanar dgraphs that are not upward planar and that admt upward straght-lne RAC drawngs[7]. Studyng the upward RAC drawablty of outerplanar dgraphs seems to be nterestng. Problem 3 Does every outerplanar acyclc dgraph admt a straght-lne upward RAC drawng? What s the tme complexty of decdng whether an outerplanar dgraph admts a straght-lne upward RAC drawng? Turnng our attenton to undrected graphs, we have shown that graphs wth degree three and sx admt RAC drawngs wth curve complexty one and two, respectvely. The followng s, however, stll open. Problem 4 What are the exact bounds for the curve complexty of RAC drawngs of bounded-degree graphs? Whle for drected graphs decdng upward straght-lne RAC drawablty s a dffcult problem, the tme complexty of decdng whether an undrected graph admts a straght-lne RAC drawng s not yet known, and consttutes, n our opnon, the man algorthmc challenge n the area.

24 76 Angeln et al. RAC Drawngs Problem 5 What s the tme complexty of decdng whether a graph admts a straght-lne RAC drawng? We have shown that there exst planar graphs that requre quadratc area n any straght-lne RAC drawng. Of course such a bound s tght for planar graphs, as planar straght-lne drawngs can be constructed n quadratc area [4, 22]. However, the followng two problems are worth studyng: Problem 6 Does every planar graph admt a RAC drawng wth curve complexty one or two n sub-quadratc area? Problem 7 What s the area requrement of straght-lne RAC drawngs of straght-lne RAC drawable graphs? Related to the last two problems, we remark that a quadratc-area lower bound for RAC drawngs (possbly wth bends) of general graphs has been provedbydgacomoet al.[8], andthattheorem0provdesacubc-arealower bound for straght-lne RAC drawngs of straght-lne RAC drawable embedded graphs. Acknowledgments Ths work started durng the Bertnoro Workshop on Graph Drawng We acknowledge Guseppe Lotta for suggestng the study of upward RAC drawngs.