A Simple 3-Approximation of Minimum Manhattan Networks

Transcription

1 A Smle 3-Aroxmaton of Mnmum Manhattan Networks Bernhard Fuchs and Anna Schulze Zentrum für Angewandte Informatk Köln, Weyertal 80, Köln, Germany and December 9, 2008 Abstract Gven a set P of n onts n the lane, a Manhattan network of P s a network that contans a rectlnear shortest ath between every ar of onts of P. A mnmum Manhattan network of P s a Manhattan network of mnmum total length. It s unknown whether t s NP-hard to construct a mnmum Manhattan network. The best aroxmatons ublshed so far are a combnatoral 3-aroxmaton algorthm n tme O(n log n), and an LP-based 2-aroxmaton algorthm. We resent a new combnatoral 3-aroxmaton for ths roblem n tme O(n log n). Both our algorthm and ts analyss are consderably smler than the ror 3-aroxmaton. Keywords: mnmum Manhattan networks, aroxmaton algorthms, rectlnear routng, VLSI desgn, 1-sanner 1 Introducton Connectng a gven set of onts wth mnmum total length s a key roblem n VLSI desgn. In the routng rocess a set of comonents, e. g., transstors or gates, have to be connected by wres. The wres are allowed to run n two erendcular drectons. The wre length sgnfcantly affects the ower consumton and the tme to sread the sgnal across the ch. Mnmum Stener trees mnmze the total wre length. Manhattan networks mose an addtonal constrant. In contrast to Stener trees they must contan a shortest ath between each ar of onts. Ths reflects the requrement to transmt sgnals between ars of comonents on the ch fast. Manhattan networks are also defned n the rectlnear metrc where one s allowed to use only horzontal and vertcal lnes. A mnmum Manhattan network s a Manhattan network of mnmum total length. See Fgure 1 (a) and for an examle. Another alcaton of Manhattan networks s descrbed by Lam et al. [PLA03], who use a varant of the Manhattan network roblem to algn gene sequences. They ask only for certan node ars to be connected by shortest aths and solve the roblem by a modfcaton of an algorthm of Gudmundsson et al. [GLN01]. Manhattan networks ft n the concet of sanners. Gven a set P of onts n the lane, a gven metrc, and a number t R wth t 1, a network s a t-sanner for P under the gven metrc, f for each ar of onts, q P, there exsts a (, q)-ath n the network of length at most t tmes the dstance between and q under the arorate norm. A Manhattan network s a 1-sanner n the rectlnear metrc. Sanners are commonly known and frst ntroduced for the Eucldean metrc by Chew [Che89]. They are studed extensvely, see for nstance the survey 1

2 (a) Q 2 Q 1 Q 3 Q 4 (c) r s (d) q Fgure 1: (a) A Manhattan network. A mnmum Manhattan network. (c) The quadrants of. (d) Global and local neghborhood. of Esten [E00] or the book of Narasmhan and Smd [NS07]. Note that the Eucldean 1-sanner s the trval comlete grah. U to now t s not known whether the mnmum Manhattan network roblem s NP-hard. Furthermore, t s not known whether a olynomal tme aroxmaton scheme exsts. Gudmundsson et al. [GLN01] ntroduced an 8-aroxmaton wth runnng tme O(n log n) and a 4-aroxmaton runnng n tme O(n 3 ). Kato et al. [KIA02] roosed a 2-aroxmaton wth runnng tme O(n 3 ), however the roof of the correctness seems to be ncomlete [BWWS06]. Cheo et al. [CNV08] resented a 2-aroxmaton based on LP-roundng. Ther LP conssts of O(n 3 ) varables and constrants. In hs PhD thess Nououa [Nou05] gave a 2-aroxmaton that runs n O(n log n). Based on the rmal-dual method, n contrast to the aroach of Cheo et al. the algorthm avods to solve an LP. Ths work s unublshed untl now. Benkert et al. [BWWS06] gave a 3-aroxmaton wth runnng tme O(n log n). Sebert and Unger [SU05] resented an aroxmaton algorthm that runs n tme O( 3 ) and clamed that t yelds a 1.5- aroxmaton. As remarked by Cheo et al. [CNV08] both the descrton of the algorthm and the erformance guarantee are somewhat ncomlete and not fully understandable. We show by a counterexamle that an mortant ntermedate ste s ncorrect, see Secton 3. We resent a 3-aroxmaton for mnmum Manhattan networks wth a runnng tme of O(n log n). Our algorthm and n some arts also the analyss s smlar to the one of Benkert et al. [BWWS06]. The lane s arttoned nto regons belongng to two dsjont areas. In the frst hase of the algorthm lne segments n the frst area are ncluded and n the second hase n the second area. For each of these two areas the aroxmaton rato s examned searately. The man dfference to the algorthm of Benkert et al. [BWWS06] les n the frst hase. Our algorthm selects lne segments n a very smle way such that both the algorthm as well as the analyss are qute easy n contrast to the aroach of Benkert et al. [BWWS06]. We wll dscuss smlartes and dfferences below. 2 Defntons We denote by x the x- and by y the y-coordnates of a ont. Each ar of onts and q sans a unque closed axs-arallel rectangle R(, q) wth and q as corners. Call R(, q) crtcal f t does not contan any other ont of P. Ths term was frst ntroduced by Sebert and Unger [SU05]. To get shortest aths t suffces to consder crtcal rectangles: If a rectangle R(, q) contans a further ont r P then t suffces to consder the two rectangles R(, r) and R(r, q). Obvously a mnmum Manhattan networks needs to contan lne segments of length at least x q x and of length y q y n the corresondng dmensons for any rectangle R(, q). For a ont R 2 we denote by Q 1 (),..., Q 4 () the four oen quadrants of. See also Fgure 1 (c). For two onts of P havng the same x- or y-coordnate each Manhattan network must contan the drect lne connectng these two onts. 2

3 b y (a) b y (c) b y b y (d) Fgure 2: The four dfferent cases of starcases. Generally, the roblem becomes easer n ths case. Snce we would have to do some rather techncally dstnctons concernng ths case, n the followng we assume the coordnates of all onts to be dfferent. We call two onts, q P (w. l. o. g. x q x and y q y ) x-neghborng f there s no further ont r wth x < r x < q x and y-neghborng f there s no further ont r wth y < r y < q y. We sometmes consder local neghborhood. More recsely, for a ont P we consder the x- or y-neghborng ont n one of ts quadrants. See for examle Fgure 1 (d). The x- neghborng ont of n the frst quadrant of s the ont q, whereas the globally x-neghborng onts of are the onts r and s. Nevertheless, f not stated otherwse we consder global neghborhood. The area defned by x- or y-neghborng onts lays an mortant role both n our algorthm and for the analyss of the aroxmaton rato. Defnton 1. For two x-neghborng onts and q the rectangle R(, q) s the rectangle R(, q) wthout the bottom and uer horzontal boundary edges of R(, q). For two y-neghborng onts and q the rectangle R(, q) s the rectangle R(, q) wthout the left and rght vertcal boundary edges of R(, q). The neghborng ont area N of a ont set P R 2 s the unon of all rectangles defned by neghborng onts. That s, N = R(, q)., q P,q x- or y-neghborng Almost all aroxmaton algorthms for mnmum Manhattan networks use starcases, but the defnton of a starcase s not standardzed. To get a clearer defnton we defne only one of four symmetrc cases of a starcase shown n Fgure 2. We defne the starcase tye as shown n Fgure 2 (a). Defnton 2. For each ont P the x-base ont s the x-neghborng ont n Q 3 (). The y-base ont b y s the y-neghborng ont n Q 3 (). Two onts belong to the same starcase sequence f they have the same base onts. The starcase sequence onts and ther base onts defne a starcase. We also consder sequences wth only one ont as a starcase sequence. To make the defnton of a starcase clear, we frst name two mortant roertes of starcases whch clarfy ther structure. Observaton 3. Each sequence ont forms a crtcal rectangle wth each base ont. Proof. Assume to the contrary that a sequence ont does not form a crtcal rectangle wth one of ts base onts. That s, the rectangle contans at least one further ont. Thus, the base ont s nether x- nor y-neghborng to the sequence ont contradctng the defnton of a starcase. 3

4 A 2 A 1 b y A 3 (a) b b (c) Fgure 3: Slttng starcases. v x b y c = v y Fgure 4: The defnton of a cross ont. In the followng we assume the sequence onts (,..., ) sorted by ncreasng x-coordnate. The next remark dslays the tycal structure of a starcase as dected n Fgure 2 (a). Remark 4. The set of onts (,..., ) wth the same set of base onts form a sequence monotone ncreasng n x- and monotone decreasng n y-drecton. Our defnton of a starcase mles that the areas A 1, A 2 and A 3 drawn n Fgure 3 (a) do not contan a ont of P. Assume to the contrary that a ont les n the area A 1. Then for at least one of the sequence onts v the x- or y-neghborng ont n Q 3 (v ) s and not as requred or b y. The starcase would slt nto at least two starcases (see Fgure 3 ). If one of the areas A 2 or A 3 contans a ont, then the starcase would slt nto at least two starcase, too (see Fgure 3 (c)). We denote the onts and of a starcase sequence (,..., ) as the outer onts of the sequence. Let and b y be the base onts of a starcase wth starcase sequence (,..., ). Let v x be the x-neghborng ont of n Q 1 ( ) and let v y be the y-neghborng ont of b y n Q 1 (b y ). Note that and v x and also and v y can be dfferent. We defne an auxlary ont c denoted as cross ont by c = (v x x, v y y). See Fgure 4. Our algorthm arttons the global Manhattan network roblem nto dsjont local Manhattan network roblems for starcases by nsertng lne segments whch searate the starcases of each other. We construct a boundary for each starcase whch s defned as follows: Defnton 5. A starcase boundary for a sequence (,..., ) of a starcase wth base onts and b y s a set of axs arallel lne segments formn a regon wth the followng roertes: For any consecutve sequence onts v and v +1, 1 k 1, the starcase boundary contans exactly one shortest ath between the two onts. Furthermore, the starcase boundary contans exactly one shortest ath between and and between and b y. The boundary of a starcase s not unque. See Fgure 5 for dfferent examles of starcase boundares for the same ont set. The ntersecton ont of the (, )-ath and the (, b y )- ath s called cross ont. 4

5 b y c b y c b y c (a) (c) Fgure 5: Dfferent starcase boundares for the same starcase r y r y b y r x b y r x (a) Fgure 6: (a) Extended base rectangles. Extended starcase boundary As requred later, we want to defne a varant of a starcase boundary whch we call extended starcase boundary. The only dfference to the revous defnton s that we do not requre the aths between outer onts and base onts to be shortest aths. Nevertheless, there have to be aths lyng n a certan area. To defne ths area we frst exand the base rectangles and call them extended base rectangles. Defnton 6. Let (,..., ) be a starcase sequence wth base onts and b y. Let r y be the y-neghborng ont n Q 2 ( ). If Q 2 ( ) P = then let r y =. Let r x be the x-neghborng ont n Q 4 ( ). If Q 4 ( ) P = then let r x =. The extended x-base rectangle s the rectangle R((x, r y y), ). The extended y-base rectangle s the rectangle R((r x x, y ), b y ). See Fgure 6 (a) for an examle of of the extended base rectangles. Now, we can defne the extended starcase boundary. Defnton 7. An extended starcase boundary for a sequence (,..., ) of a starcase wth base onts and b y s a set of axs-arallel lne segments defned n the followng way: For any consecutve sequence onts v and v +1, 1 k 1, the starcase boundary contans exactly one shortest (v, v +1 )-ath. Furthermore, the starcase boundary contans exactly one (, )-ath nsde the extended x-base rectangle and one (, b y )-ath nsde the extended y-base rectangle. See Fgure 6 for an examle of an extended starcase boundary. Snce the regon surrounded by the starcase boundary lays an mortant role, we want to name t by the followng defnton: Defnton 8. Gven a starcase S and a boundary B, the (extended) starcase area of S wth resect to B s the nteror of B. It s essental for our algorthm that after we have assgned an (extended) starcase boundary to each starcase t suffces to comute Manhattan networks for the starcases to acheve a Manhattan network for the comlete nstance. Note, that after we have comuted an (extended) 5

6 z 1,7 v 7 v 7 v 7 v 7 z 1,7 (a) (c) (d) Fgure 7: (a) and Two starcases wth the same sequence and the ndeendently comuted solutons of the algorthm of Sebert and Unger. (c) The comosed soluton. (d) A mnmum Manhattan network for the same ont set. starcase boundary for each starcase, each Manhattan network for a starcase les comletely nsde the starcase area and that each area (defned by the set of starcase boundares) s the starcase area of exactly one starcase. Thus, we slt the roblem nto a set of ndeendent subroblems of Manhattan networks for starcases. 3 The Algorthm of Sebert and Unger [SU05] In ths secton we ont out that the roof of the aroxmaton rato of the algorthm resented by Sebert and Unger [SU05] s not correct. For ths we name two dfferent onts of ther analyss, at whch t s ncorrect. Sebert and Unger frst comute two sets of vertcal and horzontal lne segments by two lane swees. The swees fx lne segments for neghborng onts such that for two neghborng onts and q there s at most one lne segment of R(, q) n the resectve dmenson fxed. A mnmum Manhattan network also contans the lengths of these lne segments, thus the two sets together contan segments of total length at most the length of a mnmum Manhattan network. They choose the cheaer of the two sets to use at most half the length of the otmal soluton. They clam that ths chosen set searates all starcases from each other and that they can ndeendently comute Manhattan networks for the starcases whch s done n the second hase of ther algorthm. In ths hase they comute for each starcase a mnmum Manhattan network contanng also a boundary for the starcase wth a dynamc rogram. For a starcase sequence (,..., ) of onts lyng to-rght whch have to be connected to a ont bottomrght of them, they select a hel ont z 1,k = (x, y ) and comute a Manhattan network for (,..., ) and z 1,k. Altogether they get a Manhattan network for the comlete nstance. Snce the chosen set of lne segments of the frst hase does not contan a comlete starcase boundary for each starcase, they also have to nsert remanng lne segments of the boundares n the second hase. They clam that ths can be done ndeendently of the other starcases (and starcase boundares) and that ths costs all n all at most the length of a mnmum Manhattan network. Ths s the crux n ther argumentaton because the boundares are not ndeendent and t s not clear how to fx ths ssue. See Fgure 7 for an examle consstng of two starcases. Frst, the algorthm of Sebert and Unger comutes for the two starcases ndeendently mnmum Manhattan networks ncludng the boundary for each starcase (Fgure 7 (a) and). The resultng network conssts of these two comuted Manhattan networks for the starcases (Fgure 7 (c)). Obvously, the otmal soluton (Fgure 7 (d)) s shorter than the soluton comosed of the two starcases comuted searately by the algorthm of Sebert and Unger n contradcton to ther clam. For ths examle the algorthm of Sebert and Unger nserts too many boundary segments. In the otmal soluton the lne segments of the boundary between the sequence onts s shared by the two starcases (see Fgure 7 (d)). 6

7 v 4 v 4 v 4 z 1,4 v 1 z 1,4 z 1,4 v 1 l z 1,4 z 1,4 v 1 l l z 1,4 v 4 v 4 v 4 (a) (c) Fgure 8: (a) The vertcal lne segment comuted by the algorthm of Sebert and Unger. The mnmum Manhattan networks for the starcases. (c) The network after nsertng the remanng connectons. After comutng Manhattan networks for starcases, they have to nsert further lne segments to establsh remanng connectons. They clam that ths can be done wthout consumng (together wth the revously comuted Manhattan networks for starcases) more than the length of a mnmum Manhattan network. See Fgure 8. Assume the set of vertcal lne segments s cheaer than the horzontal ones. They frst comute Manhattan networks for (,..., v 4 ) and z 1,4 and for (,..., v 4 ) and z 1,4 ndeendently. For the latter ont set the Manhattan network ncludes also the lne segment l. Afterwards they have to connect z 1,4 to the ont by nsertng the horzontal lne segment l. A mnmum Manhattan network for ths nstance contans only the lne segment l and not l, whch contradcts ther clam. 4 A 3-Aroxmaton of Mnmum Manhattan Networks Our 3-aroxmaton algorthm for mnmum Manhattan networks roceeds n two hases. In hase I we comute a basc set of lne segments n each of the two dmensons. These segments ensure that only sequence onts of starcases reman unconnected to the arorate base onts and that the segments consttute an extended starcase boundary wth smallest area relatve to the neghborng ont area N. More recsely, we do not need to nsert lne segments nsde N n hase II. Furthermore, we do need lne segments of a mnmum Manhattan network nsde N to justfy segments nserted by hase II because each starcase area A a comuted n hase I s contaned n a starcase area A ot (defned by lne segments nsde N ) of a mnmum Manhattan network (A a A ot ). In the second hase we comute Manhattan networks for the starcases. The general aroach to artton the roblem n a set of Manhattan network roblems for starcases s used by all combnatoral aroaches (see for examle [BWWS06], [GLN01], [KIA02] or [SU05]). We now descrbe our aroach n more detal. In hase I, we frst examne the onts from left to rght. For two x-neghborng onts and q consdered by ths swee we nsert the vertcal boundary segments of the rectangle R(, q) nto our network. Afterwards we erform an analogous swee from bottom to to. Smlarly, for two y-neghbored onts and q consdered by the swee we nsert the horzontal boundary segments of the rectangle R(, q). See Fgure 9 (a) to (c) for an examle of such a swee n the two drectons. After sortng the onts whch can be done wth runnng tme O(n log n), the swees can be erformed n tme O(n). After these two swees the only remanng art to get shortest aths between all ont ars are shortest aths between sequence onts and ther arorate cross ont. Furthermore, the u to now dentfed lne segments contan a boundary for each starcase. The aroach to consder neghborng ont ars was also consdered by Kato et al. [KIA02], Benkert et al. [BWWS06] and Sebert and Unger [SU05]. Kato et al. ntroduced the noton of 7

8 (a) (c) Fgure 9: Lne segments nserted by the swee stes. a generatng set,. e., a set of ars of onts for whch t suffces to comute shortest aths to obtan a Manhattan network for the whole nstance. Benkert et al. slt the generatng set nto two subsets for whch they comute shortest aths searately. To establsh shortest aths for all ont ars of the frst set (consstng of neghborng ont ars), they need three hases. We get these aths by roceedng only the two swees. In hase II, we comute for each starcase (gven a starcase boundary) a Manhattan network. For ths urose we dentfy all starcases. We assgn to each ont ts two base onts and then consder the set of all onts wth the same base onts as a starcase. Ths can be done n tme O(n). As stated n Secton 5 we can comute a 2-aroxmaton for a starcase n tme O(n log n) for n nut onts. Altogether, we acheve a Manhattan network for the nut onts. See Algorthm 1 for a detaled descrton. Algorthm 1 Requre: A set P R 2 of onts. Phase I: 1: Set CR =. 2: Swee over the onts of P from left to rght. Let be the currently consdered ont and q be the revously rocessed ont. Add to CR the vertcal lne segments of R(, q). 3: Swee over the onts of P bottom-u. Let be the currently consdered ont and q be the revously rocessed ont. Add to CR the horzontal lne segments of R(, q). Phase II: 4: Set MN =. 5: for each Starcase S do 6: Comute wth Algorthm 2 a Manhattan network MS of the starcase S wth the starcase boundary comuted n hase I. 7: Set MN = MN MS. 8: return MN CR. To analyze the algorthm we can nterret our strategy n such a way that we artton the lane nto regons belongng to two dsjont areas. The frst one s the neghborng ont area N, the second one the comlement R 2 \ N. In the frst hase we nclude lne segments n the frst area and n the second hase n the second area. For each of these two areas we examne the aroxmaton rato searately. Ths strategy for the analyss s smlar to the one of Benkert et al. [BWWS06]. They also consder x- or y-neghborng onts. Whereas we nclude for ars of x-neghborng onts and q both horzontal lne segments of R(, q) nto our network, they frst nclude only one of the two segments. A symmetrc statement holds for y-neghborng onts. They also get a starcase boundary for each starcase. Unfortunately, snce they nclude for two, w. l. o. g. x-neghborng, onts and q only one of the two vertcal boundary segments 8

9 r q s Fgure 10: Proof of Lemma 9. q t r s (a) Fgure 11: Proof of Lemma 10. of R(, q), these segment do not guarantee a artton nto two dsjont areas for whch they can bound the length of the nserted segments searately. Notce, that we get two dsjont areas snce we nsert both vertcal boundary segments to maxmze the area solated n the frst hase. They have to nsert further lne segments. Ths stewse roceedng comlcates the analyss and the algorthm n comarson to our aroach. To rove the correctness of Algorthm 1 we frst show that neghborng ars are connected by a shortest ath after hase I. We then show that for ars not connected after hase I t suffces to add shortest aths from sequence to cross onts. Lemma 9. After hase I of Algorthm 1 two x- or y-neghborng onts are connected by lne segments of CR formng a shortest ath. Proof. Let and q be neghborng n x-drecton. W. l. o. g. x q x and y q y. If the onts are neghborng also n y-drecton then CR contans all lne segments of R(, q). Thus, assume and q are not neghborng n y-drecton. There exsts a ont r wth y < r y < q y. W. l. o. g. let r x < x and r be the tomost one f there exst several. See Fgure 10. Thus, r s y-neghborng ether to q or to another ont s n Q 4 (q). Snce r and s are y-neghborng, the horzontal segments of R(r, s) are nserted nto CR. Together wth the vertcal segments of R(, q) they consttute a shortest ath between and q. After showng the general statement that neghborng onts are connected, we now rove that the lne segments added durng the swees of stes 2 and 3 yeld an extended starcase boundary for each starcase. That s, we get aths from the outer onts to the base onts nsde the extended base rectangles (as we rove n Lemma 10) and shortest aths between consecutve sequence onts (see Lemma 11). Lemma 10. After hase I of Algorthm 1 the outer onts of a starcase sequence are connected to both base onts by aths n the arorate extended base rectangles. There exst shortest aths between the base onts and the cross ont. Proof. Let (,..., ) be a starcase sequence wth base onts and b y. If and are x-neghborng then they are connected by a shortest ath by Lemma 9. Thus assume and are not x-neghborng. Let be the ont x-neghborng to on the rght of. The ont les above (otherwse would not be x-base ont of {,..., }). See Fgure 11. Snce s x-neghborng to we know that there exsts a shortest (, )-ath. Consder the ont q y-neghborng to above. We dstngush two cases. Frst assume q les n Q 2 ( ). (Note q 9

10 b v (a) r s v +1 b y q CP CP q q q b y Fgure 12: (a) Proof of Lemma 11. Proof of Lemma 12. s the ont r y defned n Defnton 6.) Snce s the outer ont of the starcase and s x- neghborng to, the ont q les to the left of. Agan there exsts a shortest (q, )-ath due to the neghborhood of q and. Together wth the shortest (, )-ath we get a (, )-ath n the extended x-base rectangle. See Fgure 11 (a). Now assume the ont q les to the rght of. Snce does not belong to the starcase sequence (,..., ) there s a ont above and to the left of. Let r be the one wth smallest y-coordnate. (Note r s the ont r y defned n Defnton 6.) The ont r s y- neghborng below tself to a ont s on the rght of. There exsts a shortest (r, s)-ath. Note, s not globally x-neghborng to and s an outer ont. Thus, the ont t x-neghborng to on the left of les above r. Agan there exsts a shortest (, t)-ath. Together wth the shortest (v x, )- and (r, s)-ath we get a (, )-ath nsde the extended x-base rectangle. See Fgure 11. In the same manner we can rove that we get a (, b y )-ath nsde the extended y-base rectangle. These two aths together establsh also a (, b y )- and a (, )-ath. Snce the algorthm nserts the vertcal lne segments of R(, v x ) and the horzontal segments of R(b y, v y ), we get shortest aths between the base onts and the cross ont. Lemma 11. After hase I of Algorthm 1 two consecutve onts v and v +1 of a starcase sequence are connected by a shortest ath. Proof. W. l. o. g. let v and v +1 be art of a starcase as dected n Fgure 12 (a), let v beng on the left of v +1. Assume v and v +1 are not connected after hase I. Thus, by Lemma 9 v and v +1 are nether x- nor y-neghborng. Snce v and v +1 belong to the same starcase and are consecutve n the sequence, the ont r x-neghborng to v +1 on the left s above v. See Fgure 12 (a). The vertcal segments of R(r, v +1 ) are contaned n CR. Agan, snce v and v +1 belong to the same starcase and v and v +1 are not y-neghborng there exsts a y-neghborng ont s of v below and to the rght of v +1. The horzontal segments of R(s, v ) are contaned n CR. Thus, by these four lne segments v and v +1 are connected by a shortest ath. By Lemma 10 and 11 we get an extended starcase boundary for each starcase after hase I. In the next lemma we show that t suffces to comute Manhattan networks for the starcases to get a Manhattan network for the comlete nstance. Lemma 12. If after hase I of Algorthm 1 two onts and q formng a crtcal rectangle are not connected by a shortest ath the only mssng arts to obtan such a shortest ath are connectons between and a cross ont of a starcase belongs to and between q and a cross ont of a starcase q belongs to. Proof. If and q x- or y-neghborng then there exsts a shortest ath by Lemma 9. assume that and q are nether x- nor y-neghborng. Thus, 10

11 We consder two ossble cases, namely whether and q belong to the same starcase or not. Frst, assume and q belong to the same starcase. By Lemma 11 f and q are consecutve onts of the starcase sequence then they are connected by a shortest ath. Thus, assume one ont, w. l. o. g., s a base ont of a starcase sequence q belongs to. To get a shortest ath between and q the only remanng art s a shortest ath between the cross ont and the ont q. (Note that and the cross ont are already connected by a shortest ath). Next, assume and q do not belong to the same starcase. W. l. o. g. let and q be arranged as n Fgure 12. That s, and b y are the two base onts of lyng n Q 1 () and let q and b y q be the two base onts of q n Q 3 (q). Snce s x-base ont of, the left x-neghborng ont of les below and s ether the ont or a ont below Therefore, the frst swee nserts a vertcal segment ncdent to wth end ont on the same heght or below and (by the same argument) a vertcal segment ncdent to q wth end ont on the same heght or above q. Smlarly, the second swee nserts a horzontal segment ncdent to b y wth end ont on the same wdth or on the left of and a horzontal segment ncdent to b y q wth end ont on the same wdth or on the rght of q. The cross ont of the starcase belongs to, les n the area CP marked n Fgure 12. Smlarly, the cross ont of the starcase q belongs to, les n the area CP q. By Lemma 10 there s a shortest ath between the cross ont and the arorate base onts and thus also between the two cross onts. To acheve a shortest (, q)-ath t suffces to nsert shortest aths between and q and ther resectve cross onts. Ths roves the lemma. Now we can rove that our algorthm comutes a Manhattan network. Theorem 13. For a ont set P R 2 Algorthm 1 comutes a Manhattan network. Proof. To rove that we get a Manhattan network t suffces to consder crtcal rectangles. By Lemma 12 the only onts consttutng crtcal rectangles whch are not connected after hase I are sequence onts mssng a shortest ath to an arorate cross ont. By Lemma 10 the outer onts of a sequence are connected to the base onts by aths through the cross ont after hase I of the algorthm. In hase II we comute a Manhattan network for each starcase to add mssng shortest aths between the sequence onts and the cross onts. Altogether, we obtan a Manhattan network for P. Next, we show the aroxmaton factor of the algorthm. Theorem 14. Algorthm 1 comutes a Manhattan network for P wth total length at most 3 tmes the length of a mnmum Manhattan network for P. Proof. To acheve the aroxmaton rato we have to account each length of a lne segment of a mnmum Manhattan network at most three tmes. Our roof strategy s to artton the lane nto two arts and to comare the length of a mnmum Manhattan network n each art searately. As mentoned earler ths strategy was also used by Benkert et al. [BWWS06]. Frst, recall that we only need to consder shortest aths for ont ars consttutng crtcal rectangles. The frst area we consder s the neghborng ont area N formed by the unon of all crtcal rectangles of x- or y-neghborng onts. By Lemma 12 excet for onts belongng to the same starcase all ont ars formng crtcal rectangles are connected by lne segments of CR. Consder the horzontal swee of the algorthm n ste 2. For each ar of neghborng onts and q we add the vertcal edges of the rectangle R(, q). In the mnmum Manhattan network the length of y q y s requred to connect the onts. Thus, for ths rectangle we nsert at most twce the length of the requred length. However, we must consder also neghborng crtcal rectangles havng common boundares as dected n Fgure 13. We add three lne segments to 11

12 U a L Fgure 13: Proof of Theorem 14. CR whle t suffces to nclude the mddle of the three vertcal lne segments. Generally for k onts alternatng on the lnes L and U our algorthm adds k lne segments of length a whle k/2 suffce. The rato k/ k/2 s largest for k = 3 and acheves the value 3. Ste?? works analogously. Thus, we add n hase I lne segments of length at most 3 tmes the length of segments consttutng a mnmum Manhattan network nsde the neghborng ont area N. Let and q be two neghborng nut onts. W. l. o. g. we can assume that and q are x- neghborng. We know that that there exsts a mnmum Manhattan network that adds vertcal lne segments on the boundary of R(, q) of length y q y. We add both boundary segments. Thus, t would not be reasonable to add further vertcal segments nsde R(, q). Therefore, we maxmze the area for whch we do not need to nsert vertcal lne segments after the swee (consstng of the rectangles of the neghborng ont area N defned by x-neghborng onts). At the same tme, we mnmze the remanng area for whch we have to nsert vertcal lne segments. The remanng area s the unon of the starcase areas defned by the starcase boundares gven by lne segments of hase I. Note, each starcase area s an oen area. Let (,..., v n ) be a starcase sequence wth base onts and b y and let A a be the starcase area of the boundary comuted by Algorthm 1. A mnmum Manhattan network contans shortest (, )- and (b y, v n )-aths and shortest (v, v +1 )-aths, 1 < n, nsde the neghborng ont area N consttutng a starcase boundary B ot nsde N. Let A ot be the starcase area bounded by B ot. By the constructon of the swees we get A a A ot. Furthermore, all lne segments of a mnmum Manhattan network n the nteror of the area A ot belong only to aths between sequence onts and the base onts. No such lne segment contrbutes to a shortest ath between two onts already connected by lne segments n CR. Thus, we can use the total length of a mnmum Manhattan network nsde A ot to rove the aroxmaton factor for starcases gven the starcase boundares. That s, we arttoned the roblem n two arts whch can be consdered searately. As stated above for the frst art we add at most three tmes the length of the lne segments consttutng a mnmum Manhattan network n ths area. By Theorem 16 we can comute a Manhattan network for starcases wth aroxmaton rato two. Altogether, our algorthm aroxmates mnmum Manhattan networks wth a rato of three. The core of the roof s that our hase I yelds a 3-aroxmaton outsde of starcases and mnmzes the starcase areas. That s, we get a arttonng nto two dsjont arts (defned by crtcal rectangles of neghborng onts and the starcase areas) whch can be examned searately. Thus, together wth the standard 2-aroxmaton for starcases we get a 3-aroxmaton. Theorem 15. The runnng tme of Algorthm 1 s O(n log n). Proof. Frst of all we must sort the onts of P horzontally and vertcally. Ths can be done n tme O(n log n). The swees n stes 2 and 3 then can be erformed n tme O(n). To fnd all starcases, we assgn for each ont ts two base onts. Ths can be done by a swee n tme O(n). Afterwards, we look from the ersectve of the base onts and consder all onts wth the same base onts. These onts form the starcase sequence of the starcase. Snce each ont 12

13 (a) (c) (d) Fgure 14: (a) and (c) Manhattan network comuted by our aroxmaton. and (d) Mnmum Manhattan networks for the same ont set. can be base ont of at most four dfferent starcases (each of dfferent tye), ths can be done n runnng tme O(n). The runnng tme to comute a 2-aroxmaton for a Manhattan network of a starcase gven the starcase boundary s O(k log k) for k sequence onts by Theorem 17. Each sequence ont can only belong to at most four dfferent starcases, thus we get a total runnng tme of O(n log n) to comute Manhattan networks for all starcases. Ths yelds the total runnng tme for the algorthm of O(n log n). See Fgure 14 (a) and for an examle that the aroxmaton rato of our algorthm s tght. We use three vertcal segments nstead of one. If we consder to fx ths by the observaton that we do not need the outer lne segments we see by the examle n Fgure 14 (c) and (d) that redundant lne segments can also le n the nteror of the nstance. 5 A 2-Aroxmaton of Starcases In ths secton we deal wth the roblem to comute Manhattan networks for starcases gven the starcase boundary. The algorthm arttons recursvely a starcase nto two new starcases. Ths roceedng s also known as thckest-frst arttonng and was analyzed to yeld a 2-aroxmaton by Gudmundsson et al. [GLN01] and Benkert et al. [BWWS06]. Gudmundsson et al. [GLN01] rove that gven a starcase boundary a rectangulaton of the olygon defned by the boundary wth mnmum total edge length s a mnmum Manhattan network for the onts defnng the starcase boundary. Lngas et al. show that a mnmum rectangulaton of a rectlnear olygon can be comuted n tme O(n 4 ). They state, that for a secal case of so-called hstograms ths can be comuted n O(n 3 ). They ntroduced a 2-aroxmaton for rectangulatons wth runnng tme O(n log n). We make other demands on the starcase boundary and thus we cannot use the algorthm of Gudmundsson et al. [GLN01] drectly. Furthermore, we acheve the result by comutng a Manhattan network drectly wthout the detour to the rectangulaton. The algorthm of Benkert et al. [BWWS06] to comute Manhattan networks for starcase boundares has more analoges to ours than the the one of Gudmondsson et al.. For the urose of self contanment and snce both the defntons of starcases and the boundary of starcases are not standardzed, we secfy the algorthm adated to our condtons. Gven the starcase boundary, for a ont v of the starcase sequence we denote by x the mssng lne segment of a shortest ath to the vertcal left boundary edge, by y we denote the mssng lne segment of a shortest ath to the horzontal bottom boundary edge. See Fgure 15 for an llustraton. Note that we do not count the length of the ossbly used boundary edges between v and the left and rght boundary, resectvely. Let x be the ntersecton of x wth the vertcal left boundary edge and y the ntersecton of y wth the horzontal bottom boundary edge. Our algorthm arttons the starcase nto two starcases for whch networks are comuted recursvely. In the arttonng ste two new edges are nserted, each of them beng a new boundary edge of one of the two new starcases. We get as nut an (extended) starcase boundary of a starcase wth sequence (,..., v n ) and base onts and b y. We consder the 13

14 v2 x b y }{{} c x 6 y 7 v 6 { y 1 2 Fgure 15: The defnton of x and y. segments x and y, 1 n to connect a ont v to the left and bottom boundary segment, resectvely. We select the ont v, 1 < n, for whch x y and x +1 y +1 holds, add the segments x and y +1 to our network and comute recursvely Manhattan networks for the starcases wth sequences (,..., v ) and (v +1,..., v n ). In the examle n Fgure 15 we nsert edges x 6 and y 7 and get two new starcases wth sequences (,..., v 6 ) and (v 7,..., 2 ). See Algorthm 2 RECURSION FOR STAIRCASES for a detaled descrton. The descrton assumes that the cross ont les on the orgn. Algorthm 2 RECURSION FOR STAIRCASES Requre: A starcase sequence (,..., v n ), n 3, wth boundary. 1: Set SC =. 2: Let v, v +1 be the unque ar of neghbors wth x y and x +1 y +1. 3: Set SC = SC {x } {y +1 }. 4: Let SC be the starcase recursvely comuted for the starcase sequence (,..., v ). 5: Let SC be the starcase recursvely comuted for the starcase sequence (v +1,..., v n ). 6: return SC SC SC Let (,..., v n ) be a starcase sequence wth base onts and b y. Let A a be the starcase area of the starcase boundary gven to Algorthm 2 RECURSION FOR STAIRCASES for the nstance. A mnmum Manhattan network contans shortest (, )- and (b y, v n )-aths and shortest (v, v +1 )-aths, 1 < n, nsde the neghborng ont area N consttutng a starcase boundary B ot nsde N. Let A ot the starcase area of B ot. Theorem 16. Algorthm 2 comutes a Manhattan network for a starcase wth total length at most twce the length of a mnmum Manhattan network for t f A a A ot holds. Proof. Due to the fact that we recusvely call the algorthm and n each call, we connect two onts of the sequence wth to the base, we get a Manhattan network for the starcase. We rove the rato between the length of the soluton of Algorthm 2 and the length of a mnmum Manhattan network to be two by the use of an nductve argument over the number of sequence onts. Assume we nsert n the -th ste a segment x k and a segment y k+1. Let x ot and y ot, 1 n, the segments x and y for B ot. We get x x ot and y y ot because A a A ot holds. If we nsert x then t holds x y and therefore x mn { x ot, y ot }. In an analogous manner t holds y +1 mn { x ot +1, yot +1 } f we nsert y +1. To connect v and v +1 to the cross ont a mnmum Manhattan network needs at least the length of mn { x ot + y ot +1, yot, x ot +1 }. If the mnmum s adoted for xot + y ot +1 then our algorthm 14

15 takes the rght choce. Otherwse we get the followng estmaton: x + y +1 mn { x ot, y ot } + mn { x ot +1, yot +1 } mn { x ot +1, yot } + mn { x ot +1, yot } 2 mn { y ot, x ot +1 }. So n ste k we nsert at most two tmes the requred length n the mnmum Manhattan network to connect v to the base. Furthermore, we slt the starcase n two dsjont starcase wth smaller number of sequence onts. Accordng to the nducton hyotheses for these two starcases the aroxmaton rato holds. Theorem 17. The runnng tme of Algorthm 2 s O(n log n). Proof. The only tme-consumng work to be done s ste 2 whch can be executed by bnary seach n tme O(log n). The recurson have to be roceeded O(n) tmes. Together we get the runnng tme of the algorthm to be O(n log n). 6 Acknowledgement The authors thank two anonymous referees for helful remarks and ontng out reference [Nou05]. References [BWWS06] M. Benkert, A. Wolff, F. Wdmann, and T. Shrabe, The mnmum Manhattan network roblem: Aroxmatons and exact solutons, Comutatonal Geometry: Theory and Alcatons 35 (2006), no. 3, [Che89] [CNV08] [E00] [GLN01] [KIA02] [Nou05] [NS07] L. P. Chew, There are lanar grahs almost as good as the comlete grah, Journal of Comuter and System Scences 39 (1989), no. 2, V. Cheo, K. Nououa, and Y. Vaxès, A roundng algorthm for aroxmatng mnmum Manhattan networks, Theoretcal Comuter Scence 390 (2008), no. 1, D. Esten, Sannng trees and sanners, Handbook of Comutatonal Geometry (Jörg-Rudger Sack and Jorge Urruta, eds.), Elsever, 2000, J. Gudmundsson, C. Levcooulos, and G. Narasmhan, Aroxmatng a mnmum Manhattan network, Nordc Journal of Comutng 8 (2001), no. 2, R. Kato, K. Ima, and T. Asano, An mroved algorthm for the mnmum Manhattan network roblem, Proceedngs of the 13th Internatonal Symosum on Algorthms and Comutatons (ISAAC 2002), Lecture Notes n Comuter Scence, vol. 2518, Srnger-Verlag, 2002, K. Nououa, Enveloes de Pareto et Réseaux de Manhattan: Caractérsatons et algorthmes, Ph.D. thess, Unversté de la Médterranée, G. Narasmhan and M. Smd, Geometrc sanner networks, Cambrdge Unversty Press, New York, NY, USA,

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