k th -order Voronoi Diagrams

Transcription

1 k th -order Voronoi Diagrams References: D.-T. Lee, On k-nearest neighbor Voronoi Diagrams in the plane, IEEE Transactions on Computers, Vol. 31, No. 6, pp , B. Chazelle and H. Edelsbrunner, An improved algorithm for constructing kth-order Voronoi Diagram, IEEE Transactions on Computers, Vol. 36, No.11, pp , C. Bohler, P. Cheilaris, R. Klein, C.-H. Liu, E. Papadopoulou, and M. Zavershynskyi, On the complexity of higher order abstract Voronoi diagrams, Proceedings of the 40th International Colloquium on Automata, Languages and Programming (ICALP 13), pp , Given a set S of n point sites in the Euclidean plane, the k th - order Voronoi diagram V k (S) is a planar subdivision such that each region is associated with a k-element subset H of S and denoted by VR k (H, S). all points in VR k (H, S) share the same k nearest sites H among S. V 2 (S) p {p, s} {p, q} s {r, s} {q, s} r q {q, r}

2 Property 1 Consider a Voronoi edge e between VR k (H 1, S) and VR k (H 2, S). H 1 and H 2 only differ by one site. Let H 1 \ H 2 be {p} and H 2 \ H 1 be {q}. For all points x e, H 1 H 2 are the k 1 nearest sites of x and both p and q are the k th nearest sites of x. e p x q k = 4 H 1 H 2 = 3 General Position Assumption no more than than sites are on the same line V k (S) is connected. no more than three sites are on the same circle the degree of a Voronoi vertex is exactly 3. Definition 1 Consider a Voronoi vertex v among VR k (H 1, S), VR k (H 2, S), and VR k (H 3, S). v is new if H 1 H 2 H 3 = k + 2. H 1 = H {p}, H 2 = H {q}, H 3 = H {r}, where H = k 1. the circle centered at v and touching p, q, and r will exactly enclose the k 1 sites of H. v is old if H 1 H 2 H 3 = k + 1. H 1 = H {p, q}, H 2 = H {q, r}, H 3 = H {p, r}, where H = k 2. the circle centered at v and touching p, q, and r will exactly enclose the k 2 sites of H.

3 Example p q v r v is new k = 4 H 1 = H {p} H 2 = H {q} H 3 = H {r} H = 3 p q v r v is old k = 4 H 1 = H {p, q} H 2 = H {q, r} H 3 = H {p, r} H = 2 Proprety 2 v is a Voronoi vertex among VR k (H 1, S), VR k (H 2, S), and VR k (H 3, S) (a) v is new v is an old Voronoi vertex among VR k (H 1 H 2, S), VR k (H 2 H 3, S), VR k (H 3 H 1, S). (b) v is old v belongs to VR k (H 1 H 2 H 3 ).

4 Property 3 Consider an edge e between VR k (H 1, S) and VR k (H 2, S). Then all points x e belong to VR k (H 1 H 2 ). Sketch of proof: Let H 1 \ H 2 be {p} and H 2 \ H 1 be {q}. Since e is a part of the bisector B(p, q) between p and q, the circle centered at x and touching p and q will enclose all the k 1 sites of H 1 H 2. Therefore, (H 1 H 2 ) {p, q} = H 1 H 2 are exactly the k + 1 nearest sites of x. Definition 2 For a Voronoi edge e of V k (S), if one endpoint of e is an old Voronoi vertex, e is called old; otherwise, e is called new. Property 4 New vertices of V k (S) decompose V k (S) into two kinds of connected components: 1. a new Voronoi edge 2. a connected subgraph whose internal nodes are old Voronoi vertices Each kind induces a Voronoi region of V k+1 (S). (The former comes from Property 2 (a) and Property 3, and the latter comes from Property 2(b) and Property 3.) Definition 3 For i > 1, Voronoi regions VR i (H, S) of V i (S) can be categorized into two types: type-1: VR i (H, S) contains one new edge of V i 1 (S). type-2: VR i (H, S) contains old vertices of V i 1 (S).

5 Example Type-1 p p {p, s} {p, q} s q s {q, s} q r {r, s} r {q, r} V 1 (S) V 2 (S) VR 2 ({q, s}, S) is a type-1 region because it contains one new edge of V 1 (S) Type-2 p {p, s} {p, q} q {q, s} s {p, r, s} s p {p, q, s} {p, q, r} q {r, s} r {q, r} {q, r, s} r V 2 (S) V 3 (S) Both VR 3 ({p, q, s}, S) and VR 3 ({q, r, s}, S) are type-2 regions because they contain old vertices of V 2 (S)

6 Lemma 1 For i > 1, V i 1 (S) VR i (H, S) is a tree. V i 1 (S) VR i (H, S) is V i 1 (H) VR i (H, S) Sketch of proof all points in VR i (H, S) share the same i nearest sites. V i 1 (S) partitions VR i (H, S) into at most t sub-regions, and t < i. For 1 j t, let R j be a sub-region of V i 1 (S) VR i (H, S), let H j be the (i 1)-element subset of S such that R j = VR i 1 (H j, S) VR i (H, S), and let H \ H j be {s j }. For all points x in R j, H j are the (i 1) nearest sites of x, and s j is the i th nearest site of x. In other wods, s j is the farthest site of x among H. V i 1 (S) forms the fartheset site Voronoi diagram of H inside VR i (H, S), i.e., V i 1 (S) VR i (H, S) = V i 1 (H) VR i (H, S). The farthest-site Voronoi diagram is a tree By Property 4, V i 1 (S) VR i (H, S) is a connected component, and thus V i 1 (H) VR i (H, S) is a tree. Corollary 1 If VR i (H, S) contains m old Voronoi vertices of V i 1 (S), VR i (H, S) contains 2m + 1 old Voronoi edges of V i 1 (S). Sketch of proof By the generation position assumption, the degree of a Voronoi vertex is 3. By Lemma 1, V i 1 (S) VR i (H, S) is a tree. Euler formular for a planar subdivision v e + f = 1 + c, where v is # of vertices, e is # of edges, f is # of faces, and c is # of connected component

7 Corollary 2 Under the general position assumption, E k = 3(N k 1) S k and I k = 2(N k 1) S k, where E k is # of edges, I k is # of vertices, N k is # of faces, and S k is # of unbounded faces of V k (S). Theorem 1 Given a set S of n point sites in the Euclidean plane, the total number N k of regions in V k (S) is 2k(n k) + k 2 n + 1 k 1 i=1 S i, where S i is # of unbounded regions in V i (S), and S 0 is defined to be 0. proof I i, I i respectively. are # of edges, new edges, and old edges of V i (S), respec- E i, E i tively. and I i and E i are # of vertices, new vertices, and old vertices of V i (S), N i, N i and N i are # of regions, type-1 regions, and type-2 regions of V i (S), respectively. Since an old vertex of V i+1 (S) is a new vertex of V i (S), I i+1 = I i+1 + I i+1 = I i+1 + I i I i+1 = I i+1 I i I 1 = I 1, E 1 = E 1, and E i+1 = E i+1 + E i+1 Order N i+2 type-2 regions of V i+2(s), let m j be the number of old vertices of V i+1 (S) inside the j th type-2 region of V i+2 (S), and let e j be the number of edges of V i+1 (S) inside the j th type-2 region of V i+2 (S). N i+2 j=1 m j = I i+1 = I i and N i+2 j=1 e j = E By Corollay 1, E i+1 = N i+2 j=1 e j = N i+2 j=1 i+1 (2m j + 1) = 2I i + N i+2 N i+2 = E i+1 2I i

8 N i+2 = N i+2 + N i+2 = E i+1 + (E i+1 2I i) = E i+1 2I i N 1 = n and N 2 = E 1 = E 1 = 3(n 1) S 1. since N i+2 = E i+1 2I i, E i = 3(N i 1) S i, and I i = 2(N i 1) S i, By induction on k, N k+2 = E k+1 2I k = 3(N k+1 1) S k+1 2I k k = 3(N k+1 1) S k+1 2 ( 1) k i I i = 3(N k+1 1) S k+1 2 i=1 k ( 1) k i (2(N i 1) S i ) i=1 N k = 2k(n k) + k 2 n + 1 k 1 i=1 S i Theorem 2 N k = O(k(n k)) If k n/2, by Theorem 1, N k is trivially O(k(n k)). If k > n/2, N k depends on k 1 i=1 S i Since n 1 i=1 S i = n(n 1), k 1 i=1 S i = n(n 1) n 1 i=k S i Since S i = S n i, n 1 i=k S i = n k i=1 S i N k = 2k(n k) + k 2 n + 1 k 1 i=1 S i = 2k(n k) + k 2 n + 1 n(n 1) + n 1 i=k S i = N k = 2k(n k) + k 2 n + 1 n(n 1) + n k i=1 S i Since n k i=1 S i (n k)n (recal # of k-set), N k 2k(n k) + k 2 n + 1 n(n 1) + (n k)n = k(n k) + 1

9 Iterative Construction Theorem 3 V i+1 (S) can be obtained from V i (S) by taking VR i (H, S) V 1 (S \ H) for all H S such that V i (H, S) is non-empty. Sketch of proof V 1 (S \ H) VR i (H, S) = V i+1 (S) VR i (H, S) all points in VR i (H, S) share the same i nearest sites H among S all points in VR 1 (p, S \ H) share the same nearest site p among S \ H. all points in VR 1 (p, S \ H) VR i (H, S) share the same i nearest sites H and (i + 1) st nearest site p among S, implying that VR 1 (p, S \ H) VR i (H, S) VR i+1 (H {p}, S) It is trivial that VR i+1 (H {p}, S) VR i (H, S) VR 1 (p, S \ H), VR 1 (p, S \ H) VR i (H, S) = VR i+1 (H {p}, S) VR i (H, S) for p H Corollary 3 Assume VR i (H, S) has m adjacent regions VR i (H j, S), 1 j m. Let Q be 1 j m H j \ H. Then V i+1 (S) VR i (H, S) = V 1 (Q) VR i (H, S) The proof will be an exercise. Compute V i+1 (S) from V i (S) For each nonempty region VR i (H, S), compute V 1 (Q) VR i (H, S) where VR i (H, S) has m adjacent regions VR i (H j, S), 1 j m, and Q is 1 j m H j \ H.

10 Lemma 2 V i+1 (S) can be obtained from V i (S) in O(i(n i) log n) time. Sketch of proof V 1 (Q) can be computed in Q log Q time. Q VR i (H, S) where VR i (H, S) is the boundary of VR i (H, S) O( VR i (H, S) log VR i (H, S) ) H S, H =i,vr i (H,S) = log n O( VR i (H, S) ) H S, H =i,vr i (H,S) = O(i(n i) log n) Theorem 4 V k (S) can be computed in O(k 2 n log n) time. Sketch of proof V 1 (S) can be computed in O(n log n) O(n log n) + k 1 i=1 O(i(n i) log i) = O(k2 n log n).

11 Construction by Geometric Duality and Arrangement Definition 4 (Bisectors) For two sites, p, q S, the bisector B(p, q) is {x R 2 d(x, p) = d(x, q)}. For a site p S, let B p be {B(p, q) q S \ {p}}. Definition 5 For a site p S, the k-neighborhood of p is p H,H S, H =k VR k(h, S) and denoted by VN k (p, S). VN k (p, S). Property 5 V k (S) = p S VN k (p, S) Lemma 3 VN k (p, S) is connected and each edge of VN k (p, S) is a part of the bisector B(p, q) for some q S \ {p}. The proof could be a bonus task. Lemma 4 Consider an edge of VN k (p, S). For any point x e, px intersects exactly k 1 bisectors of B p. Sketch of proof W.l.o.g, let e belong to VR k (H 1, S) VR k (H 2, S) and let p belong to H 1 \ H 2. It is clear that H 1 \ {p} are the k 1 nearest sites of x. For any q H 1 \{p}, x belongs to D(q, p), i.e., px intersects B(p, q). For any q S \ H 1, x does not belongs to D(q, p), i.e., px does not intersects B(p, q).

12 Definition 6 Given a set L of lines in the plane, let A(L) be the arrangement fromd by L. For a point x in a face of A(L), an edge e of A(L) is at level i from x if for any point y e, yx intersects exactly i 1 lines of L. The i-skeleton SK i (x, L) is the collection of edges in A(L) whose level from x is i. Lemma 5 VN k (p, S) = SK k (p, B p ) Therefore, computing V k (S) is equivalent to computing SK k (p, B p ) for all sites p S. Hereafter, we translate S such that p is located at (0, 0), and let L be B p. If we know all the vertices of SK k (p, L) and their order along SK k (p, L) (clockwise or counterclockwise, we can compute SK k (p, L) Lemma 6 Under the general position assumption, for a vertex v of SK k (p, B p ), pv intersects k 1 or k 2 lines of B p. Geometric Duality Consider a function Ψ. For a point x = (a, b) except the origin, Ψ(x) is a line : ax 1 + bx 2 = 1, and for a line l : ax 1 + bx 2 = 1, Ψ(x) is a point (a, b). Lemma 7 For an edge e of SK k (p, B p ) and any point x e, Ψ(x) partitions the plane such that one half-plane contains the origin and exactly k 1 points of Ψ(B p ). For a vertex v of SK k (p, B p ), Ψ(v) partitions the plane such that one halfplane contains the origin and k 1 or k 2 points of Ψ(B p ).

13 Example For q S \ {p}, let p q be Ψ(B(p, q)). Consider n = 8 and k = 4. p q p u p r ps l q,r l q,r corresponds to a new Voronoi vertex among VR k (H 1, S), VR k (H 2, S), and VR k (H 3, S), where H 1 = H {p}, H 2 = H {q}, H 3 = H {r}, and H = {s, t, u}. p t p q p r ps l l corresponds to a point on a Voronoi edge between VR k (H 1, S)and VR k (H 2, S), where H 1 = H {p}, H 2 = H {q}, and H = {s, t, u}. p u p t p q p u p r l q,s ps l q,s corresponds to an old Voronoi vertex among VR k (H 1, S), VR k (H 2, S), and VR k (H 3, S), where H 1 = H {p, s}, H 2 = H {q, s}, H 3 = H {p, q}, and H = {t, u}. (Note H 1 = H 1 and H 2 = H 2.) p t

14 Let v 1, v 2,... be a sequence of vertices of SK k (p, B p ) along the counterclockwise order. We consider how to compute v i+1 from v i. W.l.o.g., we let v i be the intersection between B(p, q) and B(p, r) and v i+1 be B(p, q) and B(p, s). But we do not know s. Similarly, for each q S \ {p}, let p q be Ψ(B(q, p)). Ψ(v i ) is a straight line passing through p q and p r. Let l be Ψ(v i ), and rotate l at p q in the direction such that one half-plane contains the origin and exactly k 1 points of Ψ(B p ). The rotation will hit p s first and we obtain v i+1. During the rotation, l partition Ψ(B p \ {B(p, q)})) into the same 2 sets. Property 6 Let e be an edge of SK k (p, S) and belong to B(p, q). Let v be an endpoint of e and v be an intersection between B(p, q) and B(p, s). For any point x e, let P 1 and P 2 be the 2-partition of Ψ(B p \ {B(p, q)}) formed by Ψ(x). Then, Ψ(B(p, s)) must be one of four tangent points between Ψ(B(p, q)) and the two convex hulls of P 1 and P 2. p r l p q ps p u p t

15 Lemma 8 SK k (p, B p ) can be constructed in O(n log n + SK k (p, B p ) log n) time. Sketch of proof After the sorting, it takes O(n) time to compute a vertex of SK k (p, B p ) and then view the vertex as the begining vertex v 1. It sufficient to analyze the time for computing v i+1 from v i. Assume that v i is an intersection between B(p, q) and B(p, r). Let P 1 and P 2 be the 2-partion of Ψ(B p \ {p}) formed by Ψ(v i ) and let P 1 belong to the half-plane containing the origin. If v i is a new Voronoi vertex, P 1 = k 1. let l be Ψ(v i ) rotate l at Ψ(B(p, q)) such that P 1 and Ψ(B(p, r)) belongs to different half-planes formed by l. Determine that l first touches the convex hull of P 1 or that of P 2 {Ψ(B(p, r))} Let Ψ(B(p, s)) be the first touched point of the first touched convex hull. Then v i+1 is the intersection between B(p, q) and B(p, s). Otherwise, v i is an old Voronoi vertex, and P 1 = k 2. let l be Ψ(v i ) rotate l at Ψ(B(p, q)) such that P 1 and Ψ(B(p, r)) belong to the same half-plane formed by l. Determine that l first touches the convex hull of P 1 {Ψ(B(p, r))} or that of P 2 Let Ψ(B(p, s)) be the first touched point of the first touched convex hull. Then v i+1 is the intersection between B(p, q) and B(p, s). Brodal and Jacob proposed a dynamic structure for the convex hulls allowing insertion, deletion, and tangent query in amorted O(log n) time. It takes O(n log n) time to compute the two initial convex hulls. There are O( SK k (p, B p )) insertions, deletions, and tangent queries.

16 Theorem 5 V k (S) can be computed in O(n 2 log n + k(n k) log n) time. sketch of proof V k (S) = p S SK k(p, B p ). p S O(n log n + SK k(p, B p ) log n) = O(n 2 log n + k(n k) log n)