On the N P-hardness of GRacSim drawing

Transcription

1 Journal of Graph Algorihm and Applicaion hp://jgaa.info/ vol. 0, no. 0, pp. 0 0 (0) DOI: /jgaa On he N P-hardne of GRacSim drawing and k-sefe Problem L. Grilli 1 1 Deparmen of Engineering, Univeriy of Perugia, Ialy Abrac We udy he complexiy of wo problem on imulaneou graph drawing. The fir problem, GRacSim drawing, ak for finding a imulaneou geomeric embedding of wo planar graph, haring a common ubgraph, uch ha only croing a righ angle are allowed, and every croing mu involve a privae edge of one graph and a privae edge of he oher graph. The econd problem, k-sefe, i a rericed verion of he opological imulaneou embedding wih fixed edge (SEFE) problem, for wo planar graph, in which every privae edge may receive a mo k croing, where k i a precribed poiive ineger. We how ha GRacSim drawing i N P-hard and ha k-sefe i N P-complee. The N P-hardne of boh problem i proved uing wo imilar reducion from 3-Pariion. Submied: May 2017 Reviewed: July 2017 Revied: Augu 2017 Acceped: Ocober 2017 Final: Ocober 2017 Publihed: Aricle ype: Regular paper Communicaed by: M. Beko, M. Kaufmann, F. Monecchiani Reearch uppored in par by he projec Algorimi e iemi di analii viuale di rei complee e di grandi dimenioni - Ricerca di Bae 2017, Diparimeno di Ingegneria dell Univerià degli Sudi di Perugia. A preliminary verion of hi aricle i available a arxiv: addre: luca.grilli@unipg.i (L. Grilli)

2 JGAA, 0(0) 0 0 (0) 1 1 Inroducion The problem of compuing a imulaneou embedding of wo or more graph ha been exenively explored by he graph drawing communiy. Indeed, beide i inheren heoreical inere [6], i ha everal applicaion in dynamic nework viualizaion, epecially when a viual analyi of an evolving nework i needed. Alhough many varian of hi problem have been inveigaed o far, a general formulaion for wo graph can be aed a follow: Le G 1 = (V 1, E 1 ) and G 2 = (V 2, E 2 ) be wo planar graph haring a common (or hared) ubgraph G = (V, E), where V = V 1 V 2 and E = E 1 E 2. Compue a planar drawing Γ 1 of G 1 and a planar drawing Γ 2 of G 2 uch ha he rericion o G of hee drawing are idenical. By overlapping Γ 1 and Γ 2 in uch a way ha hey perfecly coincide on G, i follow ha edge croing may only occur beween a privae edge of G 1 and a privae edge of G 2, where a privae (or excluive) edge of G i i an edge of E i \ E (i = 1, 2). Depending on he drawing model adoped for he edge, wo main varian of he imulaneou embedding problem have been propoed: opological and geomeric. The opological varian, known a Simulaneou Embedding wih Fixed Edge (or SEFE for hor), allow he edge of Γ 1 and Γ 2 o be drawn a arbirary open Jordan curve, provided ha every edge of G i repreened by he ame curve in Γ 1 and Γ 2. Inead, he geomeric varian, known a Simulaneou Geomeric Embedding (or SGE for hor), impoe ha Γ 1 and Γ 2 are wo raigh-line drawing. The SGE problem i herefore a rericed verion of SEFE, and i urned ou o be oo much rericive, i.e. here are example of pair of rucurally imple graph, uch a a pah and a ree [3], ha do no admi an SGE. Alo, eing wheher wo planar graph admi a imulaneou geomeric embedding i N P-hard [8]. Compared wih SGE, pair of graph of much broader familie alway admi a SEFE, in paricular here alway exi a SEFE when he inpu graph are a planar graph and a ree [9]. In conra, i i a long-anding open problem o deermine wheher he exience of a SEFE can be eed in polynomial ime or no, for wo planar graph; hough, he eing problem i N P-complee when generalizing SEFE o hree or more graph [13]. However, everal polynomial ime eing algorihm have been provided under differen aumpion [1,2,6,7,14,15], mo of hem involve he conneciviy or he maximum degree of he inpu graph or of heir common ubgraph. In hi paper we udy he complexiy of he Geomeric Rac Simulaneou drawing problem [4] (GRacSim drawing for hor): a rericed verion of SGE, which ak for finding a imulaneou geomeric embedding of wo planar graph, uch ha all edge croing mu occur a righ angle; of coure, analogouly o he SGE problem, every croing mu involve a privae edge of G 1 and a privae edge of G 2. We fir decribe a general N P-hardne conrucion ha ranform an inance of 3-Pariion ino a uiable inance of SEFE; ee Secion 3. Baed on hi conrucion, we how ha GRacSim drawing i N P-hard; ee Secion 4. Moreover, we inroduce a new rericed verion of he SEFE problem, called k-sefe, in which every privae edge may

3 2 L. Grilli On he N P-hardne of GRacSim drawing and of k-sefe receive a mo k croing, where k i a precribed poiive ineger. We hen how ha even k-sefe i N P-complee for any fixed poiive k (ee Secion 5), o prove he N P-hardne we ill ue a reducion from 3P, baed on he conrucion given in Secion 3. 2 Preliminarie Le G = (V, E) be a imple graph. A drawing Γ of G map each verex of V o a diinc poin in he plane and each edge of E o a imple Jordan curve connecing i end-verice. The drawing Γ i planar if no wo diinc edge inerec, excep a common end-verice. Alo, Γ i a planar raighline drawing if i i planar and all i edge are repreened by raigh-line egmen. The graph G i planar if i admi a planar drawing. A planar drawing Γ of G pariion he plane ino opologically conneced region called face. The unbounded face i called he exernal (or ouer) face; he oher face are he inernal (or inner) face. A face f i decribed by he circular ordering of verice and edge ha are encounered when walking along i boundary in clockwie direcion if f i inernal, and in counerclockwie direcion if f i exernal. A planar embedding of a planar graph G i an equivalence cla of planar drawing ha define he ame e of face for G. An ouerplanar graph i a (planar) graph ha admi a planar embedding in which all verice belong o a ame face. For a fixed poiive ineger n, a ladder L n i a graph ha can be obained by he Careian produc of a pah wih n verice and a graph coniing of a ingle edge. In oher word, L n i an ouerplanar graph wih 2n verice and n + 2(n 1) edge, which coni of wo n-verex pah, called ide pah, along wih a e of n edge, called rung, connecing he i-h verex of he fir ide pah o he i-h verex of econd ide pah (1 i n); we will ay ha a rung edge i in odd (repecively, even) poiion if i end-verice are in odd (repecively, even) poiion along heir own ide pah. For every n > 1, a ladder L n conain n 1 4-cycle, which are called he cell of he ladder; oberve ha in an ouerplanar embedding of L n no cell conain anoher. A wheel i a graph coniing of a cycle C plu a verex c and a e of edge connecing c o every verex of C; verex c i he cener of he wheel. 3 N P-hardne Conrucion The N P-hardne reul given in hi paper ue wo very imilar reducion from 3-Pariion (3P) ha rely on a ame conrucion; we refer o he following formulaion of 3P. Problem: 3-Pariion (3P) Inance: A poiive ineger B, and a mulie A = {a 1, a 2,..., a 3m } of 3m naural number wih B/4 < a i < B/2 (1 i 3m). Queion: Can A be pariioned ino m dijoin ube A 1, A 2,..., A m, uch ha each A j (1 j m) conain exacly 3 elemen of A, whoe um i B?

4 JGAA, 0(0) 0 0 (0) 3 W 1 W j W m π 1 π j π m v 0 v 1 v j 1 v j v m (a) Pumpkin gadge (b) Slice gadge Figure 1: (a) Illuraion of a pumpkin gadge (hick yle) including a baic chemaizaion of all he ranveral pah (dahed yle); we recall ha every ranveral pah coni of an alernaing equence of 2B + 1 non-hared edge. (b) Illuraion of a lice gadge encoding ineger 5. In hi ecion, we decribe hi conrucion, which ranform an inance of 3P ino an inance of SEFE, and illurae he baic idea of our reducion. We recall ha 3P i a rongly N P-hard problem [11], i.e., i remain N P-hard even if B i bounded by a polynomial in m. Alo, a rivial neceary condiion for he exience of a oluion i ha 3m i=1 a i = mb, herefore i i no rericive o conider only inance aifying hi equaliy. Le A = {a 1, a 2,..., a 3m } be an inance of 3P. We now decribe in deail a procedure o incremenally conruc an inance G 1, G 2 of SEFE aring from A; ee Fig. 1 for an illuraion of hi conrucion and Fig. 2(b) for an example of an inpu inance. A each ep, hi procedure add one or more ubgraph (gadge) o he curren pair of graph. A G 1 and G 2 have he ame verex e, for each added ubgraph we will only pecify which edge are hared and which are excluive; he final verex e will be known implicily. Hereafer, we will refer o G 1, G 2 a a full pumpkin of A. Full Pumpkin Conrucion Sar wih a biclique K 2,m+1, and denoe by, and by v 0, v 1,..., v m i verice of he parie e of cardinaliy 2 and m+1, repecively. Creae a graph G p, called pumpkin (ee, e.g., Fig. 1(a)), by adding he edge (v 0, v m ) o he biclique, i.e. G p = K 2,m+1 {(v 0, v m )}. All he edge of G p are hared edge, i.e. G p G; verice and are called he pole of he pumpkin, while (v 0, v m ) i i handle. Connec each pair of verice v j 1, v j (1 j m) of G p wih a ranveral pah π j a depiced in Fig. 1(a), which how only a baic chemaizaion of hee pah. Indeed, each ranveral pah coni of 2B + 1 non-hared edge, o ha edge in odd poiion (aring from v j 1 ) are privae edge of G 1, while hoe in even poiion are privae edge of G 2 ; hence, every ranveral pah ar and end wih an edge of G 1 and ha exacly 2B inner verice. Ineger B repreen he effecive lengh of a ranveral pah, which i defined a half he number of i inner verice. For each ineger a i A, (1 i 3m) add a ladder L 2ai+1 and aach

5 4 L. Grilli On he N P-hardne of GRacSim drawing and of k-sefe i o he pole of he pumpkin a follow: connec all he verice of one ide pah o he pole and all he verice of he oher ide pah o he pole (ee, e.g., Fig. 1(b)). The ladder L 2ai+1 and he edge connecing i ide pah o he pole of he pumpkin form a ubgraph S i called he lice gadge, which encode he ineger a i ; hence, by conrucion, he ladder of every lice alway ha an even number of cell. All he edge of S i are hared edge, excep for he rung of i ladder. In paricular, a depiced in Fig. 1(b), a rung edge in odd poiion i a privae edge of G 2, while a rung edge in even poiion i a privae edge of G 1. We conclude hi conrucion by inroducing he concep of widh w(s i ) of a lice S i : he widh w(s i ) i defined a half he number of cell in he ladder of S i, hu w(s i ) = a i. I i no difficul o ee ha a full pumpkin of A conain 6Bm + 7m + 3 verice and 8Bm + 12m + 3 edge, herefore i conrucion can be performed in polynomial ime. We oberve ha he common ubgraph i no conneced. Indeed, G coni of he pumpkin G p along wih all he edge connecing he ladder o he pole of G p and all inner verice of he ranveral pah; hu, here are 2Bm iolaed verice in he common ubgraph. Moreover, even G 1 and G 2 are no conneced, becaue in addiion o G hey alo conain heir own privae edge of lice S i (1 i 3m) and hoe of ranveral pah π j (1 j m); in paricular, due o he laer pah, G 1 and G 2 conain an induced maching of (B 1)m and Bm (privae) edge, repecively. In Secion 4 and 5, we will examine wo rericed verion of he SEFE problem, in which every SEFE drawing of a full pumpkin G 1, G 2 (if any) i alway a canonical drawing (ee, e.g., Fig. 2(b)), where a canonical drawing of G 1, G 2 i a SEFE uch ha no wo rung edge cro each oher. We now briefly explain he general raegy of our reducion. Any planar embedding of a pumpkin G p ha exacly one face conaining he pole bu no, one face conaining bu no, and m remaining face ha are inciden o boh pole. The laer face are called wedge and are ued o conain he lice gadge, which are 3m ubgraph aached o he wo pole of he pumpkin, wih no oher verice in common wih each oher and wih he pumpkin. Furher, a already menioned, every lice ha a widh ha uiably encode a diinc elemen a i of A recall ha wo diinc elemen could be equal and he rucure of a lice i ufficienly rigid (i.e. i ha a unique embedding up o a choice of he exernal face) o ha overlap and neing among lice canno occur in a canonical drawing of a full pumpkin. The baic idea of our reducion i o ge he ube A j (1 j m) of a oluion of 3P, in cae one exi, by looking a he lice in each wedge of a canonical drawing, which implie ha every wedge mu conain exacly hree lice whoe widh um o B. Of coure, wihou inroducing ome furher gadge, each wedge could conain even all lice, i.e. i widh can be conidered unlimied. Thi i he reaon why we added he ranveral pah, namely, o make all wedge of he ame widh B. Indeed, a i will be clarified laer, here canno be coing beween wo edge of a ranveral pah. Hence, he pumpkin plu he ranveral pah form a ubdiviion of a maximal planar graph, which

6 JGAA, 0(0) 0 0 (0) 5 v j 1 π j π j π j π j vj (a) Wedge W j, ranveral pah π j, and lice S j1, S j2, and S j3 v 0 v 1 v 2 v 3 (b) Canonical drawing of a full pumpkin Figure 2: (a) A wedge W j of widh 15, i ranveral pah π j, and lice S j1, S j2, and S j3 encoding ineger 4, 5 and 6, repecively. Shared edge are colored black, hoe of he pumpkin wih hick line, while privae edge of G 1 and of G 2 are colored blue and red, repecively. (b) A canonical drawing of a full pumpkin correponding o an inance of 3P wih m = 3, B = 15 and A = {4, 4, 5, 5, 5, 5, 5, 6, 6}. Slice are drawn wihin wedge according o he following oluion of 3P: A 1 = {4, 5, 6}, A 2 = {5, 5, 5} and A 3 = {6, 5, 4}. ha a unique embedding (up o a choice of he exernal face). Therefore, he effecive lengh of a ranveral pah (ha encode he ineger B) eablihe he widh of he correponding wedge. Croing beween lice and ranveral pah are hu unavoidable, becaue every ranveral pah pli i wedge ino wo par, eparaing he wo pole of he pumpkin; clearly, every lice croe only one ranveral pah. However, by uning he lengh of he ranveral pah, i.e. by chooing a uiable definiion for he effecive lengh, i i poible o form only croing ha are allowed in a canonical drawing. The key facor of he reducion i o make i poible if and only if each lice of widh a i can cro a porion of i ranveral pah wih an effecive lengh greaer han or equal o a i. In oher word, he lice rucure and he ranveral pah effecive lengh are defined in uch a way ha (i) every ranveral pah canno cro more han hree lice, and (ii) he oal widh of he lice croed by a ame ranveral pah equal ineger B, which yield a oluion of 3P. The following lemma formalize hi argumen. Lemma 1 Le A = {a 1, a 2,..., a 3m } be an inance of 3P, and le G 1, G 2 be a full pumpkin of A. G 1, G 2 admi a canonical drawing if and only if A i a Ye-inance of 3P.

7 6 L. Grilli On he N P-hardne of GRacSim drawing and of k-sefe Proof: ( ) Suppoe ha A admi a 3-pariion {A 1, A 2,..., A m }, hen a canonical drawing Γ 1, Γ 2 of G 1, G 2 can be conruced a follow. Compue a plane drawing Γ p of he pumpkin G p (ee, e.g., Fig. 1(a)) uch ha (i) he exernal face i delimied by he edge (, v 0 ), (v 0, v m ) and (v m, ) and (ii) for each j = 1, 2,..., m edge (, v j ) immediaely follow edge (, v j 1 ) in he counerclockwie edge ordering around. The drawing Γ p conain m inner face of degree four, delimied by edge (, v j 1 ), (v j 1, ), (, v j ), (v j, ) (1 j m), which are he wedge W j of he pumpkin. Conider now each riple A j = {a j1, a j2, a j3 } (1 j m), and denoe by S j1, S j2, S j3 he correponding lice in he full pumpkin. For each lice S jk (1 k 3), compue a plane drawing wih boh pole on he exernal face (ee, e.g., Fig. 1(b)). Place hee drawing one nex o he oher wihin wedge W j, in any order; for impliciy we may aume ha S j1 i he lefmo lice, S j2 i he middle lice and S j3 i he righmo one. I i no difficul o ee ha he drawing produced o far i planar, i.e. even he privae edge do no creae croing. To complee he drawing Γ 1, Γ 2, i remain o embed he ranveral pah, aking ino accoun ha every pah π j will unavoidably cro he hree lice in i wedge W j. Bu ince, by conrucion, w(w j ) = B = a j1 +a j2 +a j3 = w(s j1 )+w(s j2 )+w(s j3 ), every ranveral pah π j (1 j m) can be drawn wihin wedge W j in uch a way ha (i) every inner verex of π j i placed wihin a cell of a ladder in W j, (ii) every cell in W j conain exacly one inner verex of π j, and (iii) every croing involve wo privae edge of differen graph uch ha one i a rung edge and he oher i an edge of π j (ee, e.g., Fig. 2(a) and 2(b)). Therefore, he reuling drawing Γ 1, Γ 2 i a canonical drawing of a full pumpkin of A. ( ) We conclude he proof by howing ha if G 1, G 2 admi a canonical drawing Γ 1, Γ 2, hen A admi a 3-pariion. Le Γ p be he drawing of G p induced by Γ 1, Γ 2. Alo, le C j G p (1 j m) be he cycle coniing of he edge (, v j 1 ), (v j 1, ), (, v j ) and (v j, ). We fir claim ha he following properie are aified. (P1 ) C j (1 j m) i he boundary of a wedge W j in Γ p, where a wedge i a bounded or unbounded face of degree four in Γ p. (P2 ) Tranveral pah π j (1 j m) i drawn wihin wedge W j. (P3 ) Any wo lice canno be conained in each oher and do no overlap wih each oher excep a pole and. (P4 ) Every wedge conain exacly hree lice. Le R b (C j ) and R u (C j ) be he bounded and he unbounded plane region, repecively, delimied by C j in Γ p. Since v j 1 and v j are wo verice of C j, pah π j ha o be drawn wihin eiher R b (C j ) or R u (C j ), oherwie an inner edge of π j would cro an edge of C j, which i no allowed in a Sefe drawing of G 1, G 2 becaue C j G. Alo, if π j i conained in R b (C j ) (repecively, R u (C j )), hen all he oher pah of he pumpkin ha connec he wo pole and mu be drawn wihin R u (C j ) (repecively, R b (C j )). Properie P1 and P2 are hu aified. Concerning propery P3, i i immediae o ee ha any wo lice canno be conained in each oher. Furher, in cae of overlap, an edge e 1 of a lice S 1 would cro a boundary edge e 2 of a lice S 2, where e 2 i a privae edge of G 2 and e 1 i a privae edge of G 1. Bu hi i no poible, becaue he end-verice of e 1 are alo conneced in S 1 by a 3-edge pah coniing of wo hared edge and of a privae edge of G 2. We now how ha propery P4 i

8 JGAA, 0(0) 0 0 (0) 7 aified. We preliminarily oberve ha every lice S i (1 i 3m) mu be drawn wihin ome wedge, and all he lice in a wedge W j are croed by i ranveral pah π j. Moreover, he embedding of S i i compleely eablihed (ee, e.g., Fig. 1(b)). Indeed, ince in a canonical drawing wo rung edge canno cro each oher, i follow ha S i ha o be drawn planarly, and he wo pole and mu be on he ouer face of S i. The embedding of S i i herefore compleely deermined, becaue i i a 3-conneced planar graph and ha only one face conaining boh pole. A a conequence, no cell of S i conain anoher cell of i in Γ 1, Γ 2, i.e. he drawing of S i conain an ouerplanar drawing of i ladder. Suppoe by conradicion ha propery P4 doe no hold. Then, here would be a wedge W p (1 p m) conaining a lea four lice; recall ha here are 3m lice o be diribued among m wedge. Le u denoe uch lice by S p1, S p2,..., S pk, wih k 4, and le a pl A be he ineger encoded by lice S pl (1 l k). Since each elemen of A i ricly greaer han B/4, i follow ha k l=1 w(s pl) = k l=1 a pl > k l=1 B/4 B = w(w p), hu wedge W p i no wide enough o ho all i lice, a conradicion. In oher word, he ranveral pah π p doe no have enough inner verice o pa hrough all he cell of lice in W p avoiding croing ha are no allowed in a Sefe drawing. Now, for each wedge W j (1 j m), denoe by S j1, S j2 and S j3 he hree lice ha are wihin W j, and le a j1, a j2 and a j3 be heir correponding elemen of A. We claim ha a j1 + a j2 + a j3 = B. Indeed, i canno be 3 k=1 a jk > B, becaue i would imply ha 3 k=1 w(s jk) > w(w j ), which i no poible a een above. On he oher hand, if 3 k=1 a jk < B, here would be ome j j wih 1 j m uch ha 3 k=1 a j k > B, oherwie 3m i=1 a i would be ricly le han mb, which violae our iniial hypohei on he elemen of A. Hence, even hi cae i no poible. In concluion, every wedge W j (1 j m) conain exacly hree lice S j1, S j2 and S j3, each of hee lice ha a widh w(s jk ) (1 k 3) ha encode a diinc elemen of A, and he um of hee widh i equal o B, i.e. w(s j1 ) + w(s j2 ) + w(s j3 ) = B. Therefore, he pariioning of A defined by A 1, A 2,..., A m, where A j = {w(s j1 ), w(s j2 ), w(s j3 )}, i a oluion of 3P for he inance A. We conclude hi ecion wih wo remark. Remark 1. I i no hard o ee ha Lemma 1 remain valid if he common ubgraph G of a full pumpkin i replaced by any ubdiviion of G; we will refer o uch a modified pumpkin a a ubdivided full pumpkin. Remark 2. The previou lemma canno be uccefully applied o any Sefe drawing of a full pumpkin, becaue of he 2-edge peneraion vulnerabiliy: if rung edge can cro each oher, hen every ranveral pah π j (1 j m) can pa hrough all he cell of he ladder in W j uing only i wo fir edge; an illuraion of hi vulnerabiliy i given in Fig. 3. Alo, any enaive o pach hi vulnerabiliy by replacing he ranveral pah wih differen graph, modifying he lice accordingly, alway reuled in conrucion in which overlapping lice were poible.

9 8 L. Grilli On he N P-hardne of GRacSim drawing and of k-sefe Figure 3: Illuraion of he 2-edge peneraion vulnerabiliy. 4 NP-hardne of GRacSim drawing In hi ecion, we udy he complexiy of he following problem [4]. Problem: Geomeric Rac Simulaneou drawing (GRacSim drawing) Inance: Two planar graph G 1 = (V, E 1 ) and G 2 = (V, E 2 ) haring a common ubgraph G = (V, E) = (V, E 1 E 2 ). Queion: Are here wo planar raigh-line drawing Γ 1 and Γ 2, of G 1 and G 2, repecively, uch ha (i) every verex i mapped o he ame poin in boh drawing, and (ii) any wo croing edge e 1 and e 2, wih e 1 E 1 \ E and e 2 E 2 \ E, cro only a righ angle? Theorem 1 GRacSim drawing i N P-hard. Proof: We ue a reducion from 3P ha relie on a conrucion very imilar o ha examined in he previou ecion. In paricular, in order o ge more readable and compac GRacSim drawing, we conider a ubdivided full pumpkin G 1, G 2 of an inance A of 3P, inead of a (normal) full pumpkin. The conrucion of G 1, G 2 i a follow (ee, e.g., Fig. 4(a) and 4(b)). Conruc fir a full pumpkin of A (exacly a decribed in Secion 3). Then, ubdivide he handle edge wice, and finally, ubdivide exacly once every edge ha i inciden o one of he wo pole. A already menioned in Remark 1, hi doe no invalidae Lemma 1. Hence, o prove he aemen, i i ufficien W 1 W j W m π 1 π j π m v 0 v 1 v j 1 v j v m (a) Subdivided pumpkin (b) Subdivided lice Figure 4: (a) Illuraion of a ubdivided pumpkin gadge (hick yle) including a baic chemaizaion of all he ranveral pah (dahed yle). (b) Illuraion of a ubdivided lice.

10 JGAA, 0(0) 0 0 (0) 9 o how ha a ubdivided full pumpkin G 1, G 2 admi a canonical drawing if and only if i admi a GRacSim drawing. ( ) Suppoe ha G 1, G 2 admi a canonical drawing Γ 1, Γ 2, we how how o compue a GRacSim drawing of G 1, G 2. In Γ 1, Γ 2, every ranveral pah π j (1 j m) induce a oal order σ(π j ) on he e R j of he rung edge ha are drawn wihin wedge W j. Conruc now a pair of raigh-line drawing Γ 1, Γ 2 a follow. Draw he rung edge a a e of parallel verical egmen of he ame lengh (ee alo Fig. 5(a) and 5(b)), wih he ame baeline, and uch ha (i) he lef-o-righ order of he edge in R j coincide wih σ(π j ); (ii) for every j < j, all he edge of R j are o he lef of all he edge of R j. Le l h be he horizonal line paing hrough he midpoin of he egmen repreening he rung edge. Place verex v 0 along l h and o he lef of he lefmo edge in R 1. Similarly, place verex v m along l h and o he righ of he righmo edge in R m. Alo, place verex v j (1 j m 1) along l h, beween he righmo edge in R j 1 and he lefmo edge in R j+1. A hi poin, i i raighforward o add he remaining verice of he ubdivided pumpkin G p wihou inroducing croing. Moreover, every ranveral pah can be eaily embedded along l h in uch a way ha he e of pair of croing edge i he ame a in Γ 1, Γ 2. Therefore, he reuling drawing Γ 1, Γ 2 i a GRacSim drawing. v j 1 π j π j π j π j vj (a) Wedge W j, ranveral pah π j, and ubdivided lice S j1, S j2, and S j3 v 0 v 1 v 2 v 3 (b) GRacSim drawing of a ubdivided full pumpkin Figure 5: (a) A wedge W j of widh 15, i ranveral pah π j, and lice S j1, S j2, and S j3 encoding ineger 4, 5 and 6, repecively. (b) A GRacSim drawing of a ubdivided full pumpkin correponding o an inance of 3P wih m = 3, B = 15 and A = {4, 4, 5, 5, 5, 5, 5, 6, 6}. Slice are drawn wihin wedge according o he following oluion of 3P: A 1 = {4, 5, 6}, A 2 = {5, 5, 5} and A 3 = {6, 5, 4}.

11 10 L. Grilli On he N P-hardne of GRacSim drawing and of k-sefe ( ) Le Γ 1, Γ 2 be a GRacSim drawing of G 1, G 2. We how ha G 1, G 2 i alo a canonical drawing. Of coure, here canno be a croing beween a rung edge of a lice and a rung edge of anoher lice in Γ 1, Γ 2. Suppoe hen, by conradicion, ha wo rung edge e 1 G 1 and e 2 G 2 of a ame lice S cro each oher in Γ 1, Γ 2. Le S 2 be he ubgraph of S ha reul afer he removal of all he rung edge of G 1. I i no hard o ee ha S 2 ha a unique planar embedding E wih boh pole on he ouer face. Moreover, he end-verice of each rung edge of G 1 alway belong o a ame inner face of E. Now, le e 1 = (x, y) and le f be he face in E ha conain he end-verice x and y of e 1. If here i a croing beween e 1 and e 2 in Γ 1, Γ 2, hen e 1 ha o exi from f, croing a rung edge e 2 G 2 (poibly coinciden wih e 2 ), and he only poible way o come back o f, avoiding croing ha are no allowed in a SEFE, i o cro e 2 again. Bu hi i impoible in a raigh-line drawing. We conclude hi ecion wih wo remark. Remark 3. I i no hard o ee ha hi reducion can alo be ued o give an alernaive proof for he N P-hardne of SGE, which wa proved by Erella- Balderrama e al. [8]. Remark 4. Thi reducion canno be adaped o udy he complexiy of he one bend exenion of GRacSim, i.e. he varian of GRacSim in which one bend per edge i allowed. Indeed, in hi eing, wo rung edge can cro each oher (Lemma 1 i no longer valid) and he 2-edge peneraion vulnerabiliy canno be avoided. 5 N P-compleene of k-sefe In order o obain a more readable imulaneou embedding, which i paricularly deired in graph drawing applicaion, one may wonder wheher i i poible o compue a SEFE, where every privae edge receive a mo a limied and fixed number of croing. We recall ha here i no rericion on he number of croing ha involve a privae edge in a SEFE drawing. Furher, wo privae edge may cro more han once, and hee muliple croing could be neceary for he exience of a imulaneou embedding; however, Frai e al. [10] have hown ha whenever wo planar graph admi a SEFE, hen hey alo admi a SEFE wih a mo ixeen croing per edge pair. Moivaed by he previou conideraion, we inroduce and udy he complexiy of he following problem, named k-sefe, where k denoe a fixed bound on he number of croing per edge ha are allowed. Problem: k-sefe Inance: Two planar graph G 1 = (V, E 1 ) and G 2 = (V, E 2 ), haring a common ubgraph G = (V, E) = (V, E 1 E 2 ), and a poiive ineger k. Queion: Do G 1 and G 2 admi a SEFE uch ha every privae edge receive a mo k croing?

12 JGAA, 0(0) 0 0 (0) 11 u 1 u 2 u 3 u 4 u 5 u 0 v 0 c v 5 v 4 v 3 v 2 v 1 Figure 6: A pair of graph ha admi a k-sefe only for k 5. I i raighforward o ee ha k-sefe i, in general, a rericed verion of SEFE. Namely, for any poiive ineger k, i i eay o find pair of graph ha admi a (k + 1)-SEFE, and hu a SEFE, bu no a k-sefe. For example, conider a pair of graph G 1 = (V, E 1 ) and G 2 = (V, E 2 ) defined a follow (an illuraion for k = 4 i given in Fig. 6). The common ubgraph G = (V, E) i a wheel of 2k + 5 verice, where u 0, u 1,... u k+1, v 0, v 1,..., v k+1 are he 2(k + 2) verice of i cycle in clockwie order, E 1 = E {(u 0, v 0 )}, and E 2 = E k+1 i=1 {(u i, v k+2 i )}. Since G ha a unique planar embedding (up o a choice of he ouer face), he privae edge (u 0, v 0 ) of G 1 croe all he k + 1 privae edge of G 2, i.e. all he edge (u i, v k+2 i ) wih 1 i k + 1. Therefore, G 1 and G 2 admi a (k + 1)-SEFE, and hu a SEFE, bu no a k-sefe. Theorem 2 1-SEFE i N P-hard. Proof: We ue a reducion from 3P wih a conrucion ha i imilar, bu no idenical, o ha decribed in Secion 3. Indeed, according o he definiion of full pumpkin, every ranveral pah ha 2B + 1 inner edge, which are no enough o guaranee a mo one croing per edge; more preciely, every ranveral pah ha wo inner edge ha cro wo rung edge each (ee, e.g., Fig. 2(a)). However, hi problem can be eaily fixed by lighly increaing he lengh of a ranveral pah (modifying he definiion of effecive lengh accordingly) by adding four inner edge (ee, e.g., Fig. 7). Afer hi modificaion of he full pumpkin conrucion, he aemen can be hown wih a proof analogou o ha of Lemma 1. v j 1 π j π j π j π j vj Figure 7: Illuraion of a wedge in he conrucion for he 1-SEFE problem.

13 12 L. Grilli On he N P-hardne of GRacSim drawing and of k-sefe Theorem 3 For any fixed k 1, k-sefe i N P-complee. Proof: Concerning he N P-hardne, i uffice o repea he proof of Theorem 2, by replacing every ranveral pah π j (1 j m), wih a e of k inernally verex-dijoin pah πj 1, π2 j,..., πk j, where every πh j (1 h k) i idenical o π j. We now inroduce ome definiion and hen prove he memberhip in N P uing an approach imilar o ha decribed in [12]. An edge croing rucure χ(e 1 ) of a privae edge e 1 E 1 i a pair ε 2, σ(ε 2 ), where ε 2 i a mulie on he e E 2 \E wih cardinaliy a mo k, and σ(ε 2 ) i a permuaion of mulie ε 2. A croing rucure χ(g 1, G 2 ) of a pair of graph G 1, G 2 i an aignmen of an edge croing rucure o each privae edge of E 1. Of coure, all croing rucure of G 1, G 2 can be non-deerminiically generaed in a ime ha i polynomial in V = n, and hey include he croing rucure induced by all k- SEFE drawing of G 1, G 2. We conclude he proof by decribing a polynomial ime algorihm for eing wheher a given croing rucure χ(g 1, G 2 ) i a croing rucure induced by ome k-sefe drawing of G 1, G 2. Le G be he union graph of G 1 and G 2, i.e. G = (V, E 1 E 2 ). For each edge e of G uch ha e E 1 \ E, conider i croing rucure χ(e) = ε 2, σ(ε 2 ), replace every croing beween e and he edge in ε 2 wih a dummy verex, preerving he ordering given by σ(ε 2 ), and hen e he reuling (muli) graph for planariy. We conclude even hi ecion wih wo remark. Remark 5. Concerning he memberhip in N P, we canno uccefully apply he proof raegy ju decribed for k-sefe o GRacSim, becaue he verex coordinae of a GRacSim drawing are expreed by real number, which may have an unbounded number of decimal digi. Thu, for an arbirary inpu inance G 1, G 2, i i no poible o repreen a upere of all he GRacSim drawing of G 1, G 2 (if any), by a polynomial lengh encoding. Therefore, i remain open wheher he GRacSim drawing problem lie in N P. Remark 6. From a heoreical poin of view, i alo make ene o udy a lighly differen rericion of SEFE, where inead of limiing he number of croing per edge, i i limied he number of diinc edge ha cro a ame privae edge; recall ha wo privae edge may cro each oher more han once, which give rie o a differen problem han k-sefe. We may call hi problem k-pair-sefe, becaue k i now he bound on he allowed number of croing edge pair involving a ame edge. I i no hard o ee ha a reducion analogou o ha given in he proof of Theorem 2 and 3 can be ued o prove he N P-hardne of k-pair-sefe. The inereing heoreical apec of k-pair-sefe i he following: if k i greaer han or equal o he maximum number of edge of G i (i = 1, 2), hen a k-pair-sefe i alo a SEFE; in paricular, if k 3 V 6 he wo problem are idenical.

14 JGAA, 0(0) 0 0 (0) 13 6 Concluion and Open Problem In hi work we have hown he N P-hardne of he GRacSim drawing problem, a rericed verion of he SGE problem in which edge croing mu occur only a righ angle. Then, we have inroduced and udied he N P-compleene of he k-sefe problem, a rericed verion of he SEFE problem, where every privae edge can receive a mo k croing. Our reul raie wo main queion. Fir, a already menioned a he end of Secion 4, i would be inereing o udy he complexiy of a relaxed verion of he GRacSim drawing problem, where a precribed number of bend per edge are allowed; hi open problem wa already poed in [5]. Anoher inereing open problem i o inveigae he complexiy of k-pair- SEFE when he raio V /k end o k from he righ; we recall ha for k 3 V 6, k-pair-sefe and SEFE are he ame problem, and ha he N P-hardne of k-pair-sefe rongly relie on a conrucion where he raio V /k i ignificanly greaer han k. Acknowledgmen We would like o hank he anonymou referee for heir ueful commen and uggeion.

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