Forest-Like Abstract Voronoi Diagrams in Linear Time

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1 Foest-Like Abstact Voonoi Diagams in Linea Time Cecilia Bohle, Rolf Klein, and Chih-Hung Liu Abstact Voonoi diagams ae a well-studied data stuctue of poximity infomation, and although most cases equie Ω(n log n) constuction time, it is inteesting and useful to develop linea-time algoithms fo cetain Voonoi diagams. Fo example, the Voonoi diagam of points in convex position, and the medial axis and constained Voonoi diagam of a simple polygon ae a tee o foest stuctue and can be computed in linea time. In ode to povide a moe geneal appoach, we study abstact Voonoi diagams in a domain whee each site has a unique face touching the bounday of the domain, implying that the diagam is a foest-like stuctue, and develop a linea-time algoithm. Since abstact Voonoi diagams ae a categoy of Voonoi diagams, ou algoithm woks fo many concete Voonoi diagams. 1 Intoduction Voonoi diagams [2, 3, 9] ae a well-studied data stuctue of poximity infomation, used in many diffeent engineeing and science applications [4, 19]. In pinciple, given a set S of sites on the plane, the Voonoi diagam is a plana subdivision such that all points in a egion shae the same neaest site in S. Sites can be points, line segments, cicles, polygons, and so on, and the distance measue can be the Euclidean distance, L p noms, convex distance function, geodesic distance, and so on. Thee have been O(n log n)-time constuction algoithms fo many concete cases by andomized incemental constuction, divide-and-conque paadigm, o plane-sweep method. In ode to povide a unifying constuction method, Klein [10] intoduced Abstact Voonoi Diagams (AVDs, fo shot). Hee, no sites, cicles, o distance measues ae given. Instead, fo each pai of sites p and q fom S one takes an unbounded cuve J(p, q) = J(q, p) as pimay object, togethe with the open domains D(p, q) and D(q, p) it sepaates. Abstact Voonoi e- This wok was suppoted by the Euopean Science Foundation (ESF) in the EUROCORES collaboative eseach poject EuoGIGA/VORONOI. Chih-Hung Liu was suppoted by Alexande von Humboldt-Foundation, Gemany. Univesity of Bonn, Institute of Compute Science I, Gemany, bohle@cs.uni-bonn.de, olf.klein@uni-bonn.de, chliu 10@citi.sinica.edu.tw gions ae defined by VR(p, S) := q S\{p} and the abstact Voonoi diagam by D(p, q) V (S) := R 2 \ p S VR(p, S). The following axioms wee equied to hold fo each subset S of S. (A1) Each cuve J(p, q), whee p q, is unbounded. Afte steeogaphic pojection to the sphee, it can be completed to a closed Jodan cuve though the noth pole. (A2) Each neaest Voonoi egion VR(p, S ) is nonempty and pathwise connected. (A3) Each point of the plane belongs to the closue of a Voonoi egion VR(p, S ). The abstact Voonoi diagam can be constucted in O(n log n) steps by the divide-and-conque paadigm [10] and the andomized incemental constuction [15]. Howeve, cetain pactical applications equie only a specific substuctue of the entie diagam o a special kind of Voonoi diagam. Although the constuction time is Ω(n log n) fo many kinds of Voonoi diagams, it is still possible to compute a specific pat o a special case faste. Aggawal et al. [1] developed a lineatime algoithm fo Euclidean Voonoi diagams of points in convex position. Thei algoithm futhe allows to delete a site fom Euclidean Voonoi diagams in time linea to the stuctual changes, and also speeds up the algoithm fo the k th -ode Voonoi diagam in [17] by a O(log n) facto. Late, Klein and Lingas [12] genealized thei idea to abstact Voonoi diagams whee a Hamiltonian path passing each bisecting cuve exactly once is given called Hamiltonian abstact Voonoi diagams, and poposed a linea-time algoithm. Fo all subsets S of S such a Hamiltonian path uns though each Voonoi egion of V (S ) exactly once. Moeove, thee ae two kinds of Voonoi diagams of a simple polygon, both consisting of a tee o foest stuctue, who have eceived consideable attention. Fist, the medial axis of a simple polygon is the Voonoi

2 26 th Canadian Confeence on Computational Geomety, 2014 diagam of its polygonal edges. Lee [16] fist poposed an O(n log n)-time algoithm fo this medial axis, and Chin et al. [6] late developed a linea-time algoithm. Second, the constained Voonoi diagam of a simple polygon is the Voonoi diagam of its polygonal vetices constained by its polygonal edges. Lee and Lin [18] fist deived an O(n log n)-time algoithm fo these kinds of constained Voonoi diagams, and then Klein and Lingas [13] poposed a linea-time algoithm in the L 1 metic. Futhemoe, in the Euclidean metic, Klein and Lingas [13] late developed a andomized linea-time algoithm, and Chin and Wang [7] finally gave a deteministic linea-time algoithm. Howeve, the convex position is not applicable fo many othe geometic objects and othe distance measues. Futhemoe, computing a Hamiltonian path fo a given AVD is NP-complete, which we will pove by Coollay 4 in Section 2. Besides, those linea-time algoithms [6,7,14] fo the medial axis and the constained Voonoi diagam of a simple polygon depend on a decomposition of a simple polygon, which pevents them fom being extended to a moe geneal setting. Theefoe, we conside the abstact Voonoi diagam in a domain whee its stuctue is a foest and each of its sites has exactly one face, see Figue 1. Let D R 2 be a bounded domain, e. g. a domain bounded by, whee is a simple closed cuve intesecting each bisecto exactly twice such that no two bisectos intesect in a connected component entiely enclosed by the oute domain of. In the following, without explicit indication, V (S ) means V (S ) D and VR(p, S ) means VR(p, S ) D. p u D q t (a) V (S) s u D q t (b) V (S ) Figue 1: (a) Abstact Voonoi diagam V (S) in a domain D, the odeing of the egions along D is p, q,, s, t, u. (b) Fo a subset S S, V (S ) may be a foest, the odeing of the egions along D is hee t, q, t, s, t, u. We equie the additional axiom: (A4) V (S) is a tee. Fo all S S, V (S ) is a foest and each Voonoi egion has exactly one face. This axiom implies that each bisecto cosses D exactly once, any two elated bisectos, J(p, q) and J(p, ) having one site in common, coss at most once and, togethe with axiom (A2), D uns though each Voonoi s egion of V (S) exactly once. Based on ou axioms we pove the following esult. Theoem 1 Given a domain D togethe with the odeing of the Voonoi egions along D we can compute V (S) in time O(n). On the othe hand, if each bisecto cosses D exactly once, no two elated bisectos intesect in moe than one point and each Voonoi egion of V (S) intesects the bounday of D, meaning that no egion is empty, then we know that V (S ) is a foest fo all subsets S of S. Futhe, if we know in advance which egion of V (S) is intesected by D moe than once, we would know how to sepaate V (S) into tees and could adapt the theoem fo each tee. Othewise this would aleady be an element-uniqueness-test which would need time Ω(n log n). Thee is also a possibility of nomalizing the bisecto system in the sense that aftewads each pai of elated bisectos coss exactly once, see Section 3. Then V (S ) would be a tee fo all S S. But thee ae ( n 3) pais of elated bisectos, and none of these pais must coss fom the beginning. Thus it takes time Ω(n 3 ) to nomalize them. And even aftewads D may be a cuve intesecting each egion of V (S) exactly once but it is unclea whethe this is also tue fo subsets S of S, because only elated bisectos ae claimed to coss exactly once. That is why in this pape we chose a diffeent definition than in [12]. Compaed to this algoithm fo Hamiltonian abstact Voonoi diagams ou algoithm has two majo diffeences in the coloing (Section 4.1) and selection (Section 4.2), and we pove the coesponding theoetical popeties fo the coectness. Fo the coloing, ou algoithm needs to conside two moe subcases, and two consecutive sites in the sequence can be both coloed ed, while no consecutive sites in [12] ae coloed ed. Fo the selection, ou algoithm needs to modify V (S ) into a tee fo applying Aggawal s selecting lemma [1]. A peliminay vesion of this wok appeaed in [5]. 2 NP-completeness A Hamiltonian path with espect to an AVD is an unbounded simple cuve visiting each Voonoi egion exactly once. We show that it is NP-complete to decide whethe such a cuve exists o not. Let V (S) denote an abitay AVD. In [10], Theoem 2.7.3, it has been shown that V (S) togethe with the lage cuve aound the diagam is a biconnected plana gaph with vetex-degee 3, the vetices on ae of degee 3. Also the opposite is tue, namely each gaph fulfilling these popeties epesents an AVD. In this book a slightly diffeent definition of AVD s was used,

3 but it is easy to see that the Theoem is still tue fo ou setting based on axioms (A1) to (A3). This means that the dual of V (S), the dual of the gaph stuctue of V (S) inside, is a biconnected plana gaph and vice vesa. Thus a Hamiltonian cuve with espect to V (S) is equivalent to a Hamiltonian path, with its endpoints on the oute face, in a biconnected plana gaph. To show that it is NP-had to even decide the existence of such a path, we educe the poblem of deciding whethe a Hamiltonian cycle exits in a Delaunay tiangulation. In [8] it has been shown that this poblem is NP-complete. Lemma 2 Let G be a biconnected plana gaph. The poblem to detemine whethe a Hamiltonian path P with endpoints on the oute face of G exists is NPcomplete. Poof. It is clea that the poblem is in NP. So, let G be a Delaunay tiangulation. A Hamiltonian cycle C exists in G iff C visits each vetex v of G exactly once. Thus, iff thee exists a Hamiltonian path P with endpoints v and w such that thee is an edge fom v to w in G. Let v be an abitay vetex of G. Because G is a tiangulation, thee ae O(n) tiangles adjacent to v. Each tiangle consists of 3 vetices which ae paiwise connected by an edge. If G contains a Hamiltonian cycle, then thee must be a Hamiltonian path having its endpoints in one of the tiangles. Fo each tiangle T adjacent to v tun the gaph inside out, such that T becomes the oute face and the oute face becomes a bounded face. The esulting gaph G is biconnected and plana. Now a Hamiltonian cycle exists in G iff fo a tiangle adjacent to v a Hamiltonian path, with its endpoints on the oute face, exists in G. This poves the lemma. Fom this lemma we get ou theoem. Theoem 3 It is NP-complete to detemine if fo a given abstact Voonoi diagam thee exists an unbounded simple cuve visiting each Voonoi egion exactly once. Coollay 4 Fo a given system of bisecting cuves it is NP-comlete to detemine if thee exists an unbounded simple cuve cossing each bisecting cuve exactly once. Poof. An unbounded simple cuve cossing each bisecting cuve exactly once visits each Voonoi egion in V (S) exactly once, see Lemma 3 in [12]. 3 Nomalizing a Bisecto System Let {J(p, q) p, q S} be a system of bisecting cuves in geneal position fulfilling axioms (A1) to (A3) and a elaxed vesion of (A4): (A4 ) V (S ) is a foest fo all S S. Axiom (A4 ) implies that any two elated bisectos coss at most once. Would all pais of elated bisectos coss exactly once then V (S ) would be a tee fo all subsets S of S. Let be a closed cuve encicling all intesection points of bisecting cuves, so that each bisecting cuve consists of two unbounded segments outside of and a bounded segment inside. Notation: We shall wite p q to denote a segment of J(p, q) that has D(p, q) to its left and D(q, p) to its ight, if no confusion can aise. Theoem 5 By intoducing cossings of neighboing bisecting cuves outside of (and aftewads enlaging to include all new intesection points) we can tansfom {J(p, q) p, q S} into a nomal system {J (p, q) p, q S} of bisecting cuves such that (i) axioms (A1) to (A3) and (A4 ) ae fulfilled, (ii) each pai of elated bisectos coss exactly once, (iii) V (S) = V (S) int(). Popety (ii) is equivalent to saying that the Voonoi diagam of any thee sites contains exactly one Voonoi vetex. Namely, because of geneal position, we know that two elated bisectos J(q, p), J(p, ) can only coss in a point, whee they intesect, and J(q, ) must pass though this point, too. Such tiplet cossing points coespond to Voonoi vetices in the diagams of the thee sites involved. Fom the oiginal bisecto system, all of these ( n 3) many cossings may be missing. Poof. If the bisecto system is not nomal thee exist thee diffeent sites p, q, in S whee J(q, p), J(q, ), J(p, ) fail to coss but instead un as shown in Figue 2. By axiom (A4 ), the Voonoi egion of p in V (S) extends to infinity, w.l.o.g. though the nothen pat of the stip (the southen pat may be blocked by othe sites in V (S)). Let m denote the numbe of unbounded southen bisecto segments between q p and p. Clealy, m 1 because q is situated between q p and p. We call q p and p a stip of width m. The theoem follows by applying the following lemma epeatedly. Lemma 6 If thee is a stip of width m 1 we can intoduce anothe tiplet cossing point while maintaining popeties (1) to (3). (We obseve that the cossing point intoduced need not be the one of the stip boundaies!)

4 26 th Canadian Confeence on Computational Geomety, 2014 q p q p V R(p, S) u p q p t u q p q u q u p Figue 4: An impossible situation. m Figue 2: In pinciple we want to move q p and p togethe such that they coss on q. This may equie eodeing the bisecto segments in between, as we must not cause elated bisectos to coss moe than once. q p t p t q q Figue 3: Illustation of the case t u = t q. Poof. By induction on m. If m = 1 then q p, q, p ae diect neighbos and can be baided to obtain a tiplet cossing point, v. We enlage to include v, and have obtained the new odeing p, q, q p. Axioms (A1) to (A4 ) ae still fulfilled. Now let m > 1, and let t u be the ight hand side neighbo of q p, as shown in Figue 2. If t u = q then we ae done with moving q p to the ight, and stat to move p to the left towads q in a symmetic way. If t, u ae diffeent fom q, p we can simply make q p coss t u without difficulty. This educes m to m 1, and the claim follows by induction. Othewise, we analyze the following cases. t u = p u. Impossible, because q u must appea between q p and p u. t u = t p. Hee t p and p fom a stip of width m 1, so that induction applies. t u = t q. We obseve that t p because q p uns to the noth. Similaly, we have t. Since the egion of q is nonempty, q p and t q must intesect, as shown in Figue 3. But then t p must appea in between a contadiction. t u = q u. This case splits into thee subcases, depending on the intesection behavio of q u. Fist, if q p and q u coss we have the situation shown in Figue 4. We obseve that q u cannot coss q, too, because it would need to coss it twice in ode to un to q p. Since the egion of p is unbounded to the noth, u p and p cannot coss, so u must un between them to the noth. If u wee situated, on the southen pat of, between q p and q u, as shown in Figue 4, it could not un to the noth because it could not coss q u (the cossing would have to lie on q, too, which is impossible as q u and q ae disjoint). Thus, u appeas between q u and p, and q u and u fom a stip of width < m. Induction applies because the egion of u is unbounded to the noth. Second, let us assume that q u intesects neithe q p no q, as shown in Figue 5. In the Voonoi diagam of p, q, u, bisecto q u must un between q p and p u without cossing, because the egion of p uns to the noth. If, on the southen bounday of, bisecto p u appeas to the left of p then q p and p u fom a stip of width < m, and induction applies. Othewise, we have the situation depicted in Figue 5. Since q u and q ae disjoint by assumption, q and u cannot coss. Thus, q u and u fom a stip of width < m, and we can apply induction since the egion of u is unbounded to the noth. Thid, we assume that q u intesects q but not q p; see Figue 6. As in the pevious case, bisecto q u must un between q p and p u without cossing, and if p u appeas to the left of p we can apply induction to the stip fomed by q p and p u. Let us assume that p u is situated to the ight of p. If p and p u wee disjoint, eithe the egion of o of u would be empty in the Voonoi diagam of p,, u, because u would un to the left of p ; see Figue 7. Thus, thee must be a cossing, as shown in Figue 6. But thee is no way the two segments of u can be connected, since multiple cossings ae not allowed a

5 contadiction. This concludes the poof of the Lemma and of the Theoem. p q p q u q u p u Figue 5: Bisectos q u and u fom a stip of width < m. q p q u q u u p p u Figue 6: An impossible situation, since the two segments of J(, u) cannot be connected. u p p u Figue 7: The Voonoi egion of o of u would be empty. 4 The Algoithm Let us get back to ou actual poblem. Hee a domain D togethe with the odeing of the Voonoi egions along its bounday is given. Definition 1 Fo each set of sites S S let π(s ) be the sequence of egions of V (S ) along D. Since V (S) patitions D into π(s) pieces, each element of π(s) coesponds to a unique piece. Fo each element p of π(s), d(p) is a point on its coesponding piece. Remak that VR(p, S) VR(p, S ), thus d(p) VR(p, S ) fo all subsets S of S. Actually, π(s) depends on the stating point and the diection of a tavesal along D. W. l. o. g., we assume the stating point is known and the diection is clockwise. Axiom (A4) implies that each element in π(s) occus only once. Fo subsets S of S we have the following obsevation. Lemma 7 π(s ) is a Davenpot-Schinzel-Sequence of ode 2. Poof. By definition no element of the sequence appeas twice without anothe site in between. So, suppose thee ae p q S such that p, q, p, q occu in this odeing in π(s ). Then eithe the two p s o the two q s can not be connected in D, a contadiction to axiom (A4). We want to use a ecusive algoithm to compute V (S). To be able to ecusively compute V (S ) fom V (S) it is impotant that the input, the sequence of sites π(s ), fulfills the same popeties as the sequence π(s). But π(s) is a Davenpot-Schinzel-Sequence (DSS) of ode 1, wheeas π(s ) may be a DSS of ode 2. Fo this pupose we will use the following definition. Definition 2 Let π (S ) be the subsequence of π(s) containing all elements fom S, i. e. π (S ) is a DSS of ode 1. In the following we show that it indeed suffices to conside the subsequence π (S ) in ode to compute V (S ). Now ou algoithm can be summaized as follows: 1. Colo each element of π(s) eithe blue o ed, i.e., π is patitioned into π (B) and π (R), and S is patitioned into B and R, such that both B and R ae a constant faction of S, and fo each two consecutive ed sites, 1, and 2, in π, VR( 1, B { 1, 2 }) and VR( 2, B { 1, 2 }) ae not adjacent. See Section 4.1 fo details.

6 26 th Canadian Confeence on Computational Geomety, Compute V (B) fom π (B) ecusively. 3. Select a subset C fom R such that C is a constant faction of R, and fo any two sites, c 1 and c 2, VR(c 1, B {c 1, c 2 }) and VR(c 2, B {c 1, c 2 }) ae not adjacent. See Section 4.2 fo details. l l m m p p q q Add atificial Voonoi edges to obtain a tee stuctue V (B) Apply Aggawal et al. s selecting Lemma [1] on V (B) 4. Compute V (B C) by sequentially inseting each element of C into V (B). 5. Compute V (G) fom π (G) ecusively, whee G = R\C and π (G) is obtained fom π (R) by emoving all elements in C. 6. Mege V (B C) and V (G). Step 1 can be caied out in linea time accoding to Section 4.1, Step 3 and Step 4 can be completed in linea time accoding to Section 4.2 and 4.4, and Step 6 can be implemented in linea time using the geneal mege method descibed in [10]. Since B and G is a constant faction of S, the above claims conclude Theoem 1. Definition 3 Fo a set S of sites, a subset S of S, and a site p of S, a connected intesection between VR(p, S ) and D is edundant if it does not contain the connected intesection between VR(p, S) and D. Fom the viewpoint of π (S ), which is a subsequence of π(s), a connected intesection between VR(p, S ) and D is edundant if it does not contain d(p) fo p in π (S ). Definition 4 Fo all S S, a pq-vetex of V (S ) is a Voonoi vetex adjacent to VR(p, S ), VR(q, S ), and VR(, S ) clockwise. If VR(p, S ) is the only egion bodeing VR(q, S ), VR(p, S ) encloses VR(q, S ), fo bevity we say p encloses q in V (S ). 4.1 Red-Blue-Coloing Scheme The coloing scheme consists of two steps, whee a site is blue as long as it is not coloed ed. See also Figue Fo evey 5 consecutive sites along π(s), (l, m, p, q, ), p is coloed ed if one of the following conditions holds. Let T be {l, m, p, q, }. (i) Thee is a mpq-vetex in V (T ). (ii) VR(m, T ) encloses VR(p, T ). (iii) VR(q, T ) encloses VR(p, T ). 2. Fo evey 3 consecutive sites along π(s) that ae all blue, the middle one is coloed ed. Figue 8: Two cases whee p is coloed ed. Let R be the set of ed sites, and B be the set of blue sites. Obseve that the final diagam V (S) is a tee, but in the ecusion V (S) may be a foest, e. g. when S = B. Then we use the sequence π (S) instead of π(s). Lemma 8 No 3 consecutive sites in π(s) ae all coloed ed. Poof. Fo the sake of a contadiction assume that thee consecutive sites 1, 2, 3 ae all ed. Let s 1 and s 2 be the two consecutive sites pevious to 1, and s 3 and s 4 the two consecutive sites afte 3. By definition 1, 2 and 3 can not be coloed ed by step 2. Thus we need only conside step 1. Thee ae thee diffeent cases fo 1 to be coloed ed. Case 1: Thee is an s vetex in V ({s 1, s 2, 1, 2, 3 }). This vetex is still an s vetex in V ({s 2, 1, 2, 3 }) implying that thee can not exist an vetex in V ({s 2, 1, 2, 3 }) and hence also no vetex in V ({s 2, 1, 2, 3, s 3 }). This means that 2 must be coloed ed because 1 o 3 encloses it in V ({s 2, 1, 2, 3, s 3 }). But then 2 can not be adjacent to a vetex in V ({s 2, 1, 2, 3 }), i.e., no s vetex exists, a contadiction. Case 2: 1 is coloed ed because s 2 encloses 1 in V ({s 1, s 2, 1, 2, 3 }). But then the egions of 1 and 2 ae not adjacent in V ({s 2, 1, 2, 3 }) and thee can be no vetex in V ({s 2, 1, 2, 3, s 3 }). Futhe 1 can not enclose 2. This means that 2 must be coloed ed because 3 encloses it in V ({s 2, 1, 2, 3, s 3 }). But then thee can be no 2 3 s 3 - vetex in V ({ 1, 2, 3, s 3, s 4 }) and 3 can not be enclosed by 2 o s 3 in V ({ 1, 2, 3, s 3, s 4 }). Thus 3 is not coloed ed, a contadiction. Case 3: 1 is enclosed by 2 in V ({s 1, s 2, 1, 2, 3 }) but then because of the same easons as in case two 2 is not coloed ed. Coollay 9 Let s 1, s 2, 1, 2, s 3, s 4 be 6 consecutive sites in π(s). If 1 and 2 ae both ed, then s 2 and s 3 ae both blue. Futhe s 2 encloses 1 in V ({s 1, s 2, 1, 2, s 3 }) and s 3 encloses 2 in V ({s 2, 1, 2, s 3, s 4 }). In paticula s 2 encloses 1 and s 3 encloses 2 in V ({s 2, 1, 2, s 3 }). Poof. Lemma 8 shows that s 2 and s 3 ae both ed. The case two in the poof of Lemma 8 is the only case

7 whee two consecutive sites 1 and 2 ae both coloed ed. Hee s 2 encloses 1 in V ({s 1, s 2, 1, 2, s 3 }) and 3 encloses 2 in V ({s 2, 1, 2, s 3, s 4 }), implying that s 2 encloses 1 and s 3 enclose 2 in V ({s 2, 1, 2, s 3 }). It can happen that two consecutive sites ae both coloed ed, see Figue 9. Howeve, we still have the following popety. s 1 s π (S) = (s 1, s 2, 1, 2, s 3, s 4) π(s) = (s 1, s 2, 1, s 2, s 3, 2, s 3, s 4) s 3 s 4 Figue 9: Two consecutive sites 1 and 2 in π (S) ae both coloed ed, but thei egions ae not adjacent Lemma 10 Let 1 and 2 be two consecutive ed sites. Then VR( 1, B { 1, 2 }) and VR( 2, B { 1, 2 }) ae not adjacent. Poof. Let s 1 be the site pevious to 1 and s 2 the site afte 2 in π. Thee ae thee cases. Case 1: Thee is no blue site between 1 and 2. Because of Coollay 9, s 1 and s 2 ae both blue and s 1 encloses 1 and s 2 encloses 2 in V (s 1, 1, 2, s 2 ). Thus it follows diectly that the egions of 1 and 2 can not be adjacent in V (B { 1, 2 }). Case 2: Thee is exactly one blue site b between 1 and 2. Fo the sake of a contadiction suppose VR( 1, B { 1, 2 }) and VR( 2, B { 1, 2 }) ae adjacent. Then the egions of 1 and 2 ae the only egions that may be adjacent to the egion of b in V (B { 1, 2 }). If they both ae adjacent to the egion of b, then thee is a 1 b 2 -vetex in V (B { 1, 2 }). If only the egion of 1 is adjacent to the egion of b, then 1 encloses b in V (B { 1, 2 }) and if only the egion of 2 is adjacent to the egion of b, then 2 encloses b in V (B { 1, 2 }). Now if s 1 and s 2 ae both blue, then {s 1, 1, b, 2, s 2 } B { 1, 2 } and b would have been coloed ed, a contadiction to the assumption that b is blue. Now assume s 1 is ed and let s 0 be the pedecesso of s 1. Coollay 9 tells us that b encloses 1 in V ({s 0, s 1, 1, b, 2 }), but then the egions of 1 and 2 can not be adjacent in V (B { 1, 2 }). The case that s 2 is ed is symmetic. Case 3: Thee ae exactly two blue sites b 1 and b 2 between 1 and 2. As in case 2, fo the sake of a contadiction suppose VR( 1, B { 1, 2 }) and VR( 2, B { 1, 2 }) ae adjacent. Then the egions of 1, 2, b 1 and b 2 ae the only egions that may be adjacent to the egions of b 1 and b 2 in V (B { 1, 2 }). Now thee ae two subcases: Case 3.1. The egion of b 1 o b 2 is not adjacent to any of the egions of 1 o 2 in V (B { 1, 2 }). Then b 1 encloses b 2 o b 2 encloses b 1 in V (B { 1, 2 }). W.l.o.g. let b 1 enclose b 2, the othe case is symmetic. If s 2 is blue, then b 1 also encloses b 2 in V ({ 1, b 1, b 2, 2, s 2 }), and b 2 must be coloed ed. But if s 2 is ed, then by Coollay 9 b 2 has to enclose 2 in V ({b 1, b 2, 2, s 2 )}), both a contadiction. Case 3.2. Both the egions of b 1 and b 2 ae adjacent to the egion of 1 o 2. Because all Voonoi egions ae connected the egion of b 1 o b 2 is adjacent to the egion of 1 but not 2 o vice vesa. Let b 1 be adjacent to the egion of only 1 and b 2, the othe 3 constellations ae symmetic. Then thee is a 1 b 1 b 2 -vetex v in V (B {b 1, b 2 }). If s 1 is blue, then v is also a 1 b 1 b 2 - vetex in V ({s 1, 1, b 1, b 2, 2 }) and thus b 1 would be ed, a contadiction. If s 1 is ed, then 1 would be enclosed by b 1 in V ({s 0, s 1, 1, b 1, b 2 }), a contadiction as in Case Choosing Cimson Sites We want to apply the following combinatoial lemma fom [1] to obtain an independent set of cimson sites. Lemma 11 Let T be a binay tee embedded in the plane and fo each leaf l a subtee T l ooted at l is defined. Futhe, the subtees of two consecutive leaves in the topological odeing aound T ae disjoint. Then one can in linea time find a fixed faction of leaves whose subtees ae paiwise disjoint. To use this lemma we modify the foest V (B), geneated by the blue sites, by adding some edges and leaves to obtain a tee V (B) fulfilling the claimed popeties. We stat with the following obsevation. Lemma 12 We can detect all edundant intesections of V (B) in time O(n). Poof. Fist compute π(b) by deleting all sites fom R fom π(s). This takes time O(n). Recall that π(b) is the sequence of sites along D in V (B). This is a Davenpot-Schinzel-Sequence of ode 2, wheeas π (B) is a Davenpot-Schinzel-Sequence of ode 1. Let B = m n, π (B) = (p 1,..., p m ) and π(b) = (q 1,..., q l ), whee q 1 = p 1 efes to a non edundant intesection. Futhe l 2m 1, because π(b) is a DSS of ode 2. Let q ij = p j efe to a non edundant intesection and let q ij+1 be the fist p j+1 afte q ij. We claim that q ij+1 = p j+1 efes to a non edundant intesection implying that all q between q ij and q ij+1 efe to edundant intesections. Suppose q ij+1 = p j+1 is edundant. Then thee must be a q = p j+1 afte q ij+1 in π(b) efeing to the non edundant intesection of p j+1. This means that all q

8 26 th Canadian Confeence on Computational Geomety, 2014 between q ij+1 and q ae edundant and thus fo all such q p j+1 thee is anothe q befoe q ij+1 o afte q in π(b). Because all faces of V (B) D ae connected in D, the intesection of q ij+1 can be connected to the intesection of q by path in D VR(p j+1, B). But then no q p j+1, thee must be at least one between q ij+1 and q, can be connected to any intesection of q befoe q ij+1 o afte q by a path in D VR(q, B), a contadiction. Now we constuct V (B) out of V (B) by the following opeations, compae Figue 10. (i) Fo all edundant intesections on D link the two leaves bounding it along D. If the edundant intesection bodes anothe edundant intesection on its ight end, then let the leaf between them now be a vetex in V (B). Obseve that this is a vetex of degee 3. Othewise connect the ight end of the link to V (B) without ceating a vetex. The same is done on the left side of the edundant intesection. Next we attach some leaves to V (B) outside of D such that between each pai of consecutive blue sites b i and b i+1 having one (o two) ed site(s) in between, thee is exactly one (o between one and two) leaves in V (B). If thee is no ed site between b i and b i+1 thee is also no leaf. (ii) If thee ae one o two ed sites 1 and 2 between two consecutive blue sites b i and b i+1 but no leaf between them, then thee is a connected set of edundant intesections between b i and b i+1. If d( j ), j = 1, 2 lies within this sequence we attach a leaf to V (B) at d( j ), othewise if d( j ) lies to the left (ight) of the sequence we attach a leaf at the leftmost (ightmost) point of the sequence. If both d( j ), j = 1, 2 ae to the left (ight) of the edundant intesection sequence, only one leaf is attached at the leftmost (ightmost) point. Between two consecutive blue sites thee ae at most two ed site. Thus fo evey connected sequence of edundant intesections at most two leaves ae attached. Futhe, between each pai of consecutive blue sites sepaated by one o two ed sites thee is now at least one leaf. (iii) If thee is a leaf in V (B) between two consecutive blue sites b i and b i+1, which ae not sepaated by a ed site, it is puned like in [12]. Lemma 13 V (B) is a binay tee and can be constucted in time O(n). d(k) l p π(s) = (k, l, m, p, q,, s, t, u, v, w, x, y...) π (B) = (l, p,, t, u, x, y...) π(b) = (t, l, p,, t, u, t, y, x, y,...) t u d(v) d(w) x y Figue 10: V (B), fat edges indicate edundant intesections and new leaves. Poof. It is clea that V (B) is a foest. So assume it is disconnected. Then thee is a site b B whose Voonoi egion in V (B) intesects D in moe than one component. By (i) all these components ae non edundant. But then b has to appea seveal times in π(s), a contadiction. Definitions (i) to (iii) imply that all intenal nodes of V (B) ae of degee 3. By lemma 12 we can detect all edundant intesections in time O(n). In the same time opeation (i) can be accomplished. Fo opeation (ii) and (iii) we have to walk once aound D and look at consecutive blue sites. Fo each pai of consecutive blue sites b i and b i+1 we test if thee ae zeo, one o two ed sites in between. If thee is no ed site between b i and b i+1 but a leaf, we pune the leaf in constant time. If thee ae one o two ed sites 1 and 2 but no leaf between b i and b i+1 we test if d( 1 ) and d( 2 ) lie to the left, within o to the ight of the edundant intesection sequence between the two blue sites and attach one o two leaves like descibed in (ii). Fo each edundant intesection this takes constant time and thee ae O(n) edundant intesections altogethe. 4.3 Coloing Cimson If two blue sites b i and b i+1 ae sepaated by a ed site in π(s) but the leaf between them is not contained in VR(, B {}), then is enclosed by the egion of b i (o b i+1 ) in V (B {}). In this case colo cimson with espect to b i (b i+1 ) and if the leftmost (ightmost) leaf between b i and b i+1 is not contained in the egion of a consecutive site, associate with the subtee containing only this leaf. If two ed sites ae between b i and b i+1 and both ae coloed cimson because of b i (b i+1 ) associate only one of them with the leftmost (ightmost) leaf. Up to now we may aleady have coloed some ed sites cimson. To make sue we eceive a fixed faction of cimson sites we apply lemma 11 in the following way. Fo each leaf l of V (B) contained in a ed egion VR(, B {}) define T l by the subtee spanned by all vetices of V (B) contained in VR(, B {}). The next lemma shows that this is possible.

9 Lemma 14 Let be a ed site. If VR(, B {}) intesects a leaf of V (B), then VR(, B {}) V (B) is connected. Othewise it is empty. Poof. Suppose VR(, B {}) intesects a leaf of V (B) and VR(, B {}) V (B) is not connected. Then VR(, B {}) would disconnect the egion of a blue site in V (B {}), a contadiction. If VR(, B {}) does not intesect a leaf of V (B) it must be contained within a single blue egion of V (B), thus it can not intesect V (B). Now Lemma 10 and 11 imply the equested popety. Lemma 15 No egions of two cimson sites ae adjacent in V (B C). Poof. To test whethe the paent node of a leave is contained in the egion of a ed site it is enough to conside the diagam of the thee sites adjacent to the node and the ed site. Thus this test can be done in constant time. Futhe each leaf of F is associated with the egion of a ed site. This ensues a coect application of Lemma 11 and finishes the poof. 4.4 Insetion of Cimson Sites We can now inset the cimson sites into V (B) in ode to eceive V (B C). Fo each cimson site c whose egion does not intesect a leaf of V (B) we know that it is enclosed by the egion of a blue site b i o b i+1. Let it be b i, then we just have to inset the pat of the bisecto J(, b i ) contained in VR(b i, B) as a new edge in V (B C). The othe cimson sites can be inseted along the subtees of V (B) associated with them. Thus also the insetion takes time O(n). 5 Discussion A natual question is if it is possible to elax axiom (A4) and still have a linea time algoithm fo computing the Voonoi diagam. Thee exist applications whee Voonoi egions esticted to the domain D ae disconnected. This is something that can happen fo the fathest Voonoi diagam of line segments o when the domain D coesponds to a Voonoi egion which is to be deleted fom a given Voonoi diagam. Refeences [1] A. Aggawal, L. J. Guibas, J. Saxe, and P. W. Sho. A linea-time algoithm fo computing the Voonoi diagam of a convex polygon. Discete and Computational Geomety 4, pp , [2] F. Auenhamme. Voonoi Diagams: A Suvey of a Fundamental Geometic Data Stuctue. ACM Computing Suveys 23(3), pp , [3] F. Auenhamme and R. Klein. Voonoi Diagams. In: J.R. Sack and G. Uutia (Eds.), Handbook on Computational Geomety, Elsevie, pp , [4] F. Auenhamme, R. Klein, and D.-T. Lee. Voonoi Diagams and Delaunay Tiangulations. Wold Scientific Publishing Company [5] C. Bohle, R. Klein, and C.-H. Liu. Foest-Like Abstact Voonoi Diagams in Linea Time. Poc. 30th Euopean Wokshop on Computational Geomety [6] F. Chin, J. Snoeyink, and C. A. Wang. Finding the medial axis of a simple polygon in linea time. Discete Computational Geomety 21, pp , [7] F. Chin and C. A. Wang. Finding the constained Delaunay tiangulation and constained Voonoi diagam of a simple polygon in linea time. SIAM Jounal on Computing 28(2), pp , [8] M. Dillencout. Finding Hamiltonian Cycles in Delaunay tiangulations is NP-complete. Discete Applied Mathematics 64, pp , [9] S. Fotune. Voonoi diagams and Delaunay tiangulations. In: J.E. Goodman and J. O Rouke (Eds.), Handbook of Discete and Computational Geomety, Chapte 20, CRC Pess LLC, pp , [10] R. Klein. Concete and Abstact Voonoi Diagams. Lectue Notes in Compute Science 400, Spinge- Velag, [11] R. Klein, E. Langetepe, and Z. Nilfooushan. Abstact Voonoi Diagams Revisited. Computational Geomety: Theoy and Applications 42(9), pp , [12] R. Klein and A. Lingas. Hamiltonian abstact Voonoi diagams in linea time Intenational Symposium on Algoithms and Computation (ISAAC 94), pp , [13] R. Klein and A. Lingas. Manhattonian poximity in a simple polygon. Intenational Jounal of Computational Geomety and Applications 5, pp , [14] R. Klein and A. Lingas. A linea-time andomized algoithm fo the bounded Voonoi diagam of a simple polygon. Intenational Jounal of Computational Geomety and Applications 6(3), pp , [15] R. Klein, K. Mehlhon, and St. Meise. Randomized incemental constuction of abstact Voonoi diagams. Computational Geomety 3, pp , [16] D. T. Lee. Medial axis tansfomation of a plana shape. IEEE Tansaction on Patten Analysis and Machine Intelligence 4(4), pp , [17] D. T. Lee. On k-neaest Neighbo Voonoi Diagams in the Plane. IEEE Tans. Computes 31(6), pp , [18] D. T. Lee and A. Lin, Genealized Delaunay tiangulations fo plana gaphs, Discete Computational Geomety 1, pp , 1986 [19] A. Okabe, B. Boots, K. Sugihaa, and S. N. Chiu. Spatial Tessellations: Concepts and Applications of Voonoi Diagams. Wiley Seies in Pobability and Statistics, 2000.