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2. Voronoi Diagram 2.1 Definiton Given a finite set S of points in the plane , each point X of defines a subset S X of S consisting of the points of S closest to X. These subsets S X define further an equivalence relation: X Y iff S X = S Y A B X Y Example: S X = {A} S Y ={A,B} = S Z Z C D E

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The equivalence classes of this relation: points, line segment segments (including half lines), and polygonal regions (including unbounded), are the faces of the corresponding lattice called Voronoi Diagram of S. The cell {X: S X = {A}} will be denoted C(A). We will see that Voronoi Diagram of S is closely related to Delaunay Triangulation of S as well as to the convex hull of its projection onto some paraboloid. In both cases the connection is made with the use of duality. Voronoi Diagram (cont.)

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Equivalence in E 2 2.2 Duality in E 2 / E 3 Voronoi Diagram (duality cont.) Set A of vertices, edges and polygons is combinatorially equivalent to such a set B if there is a 1-1 mapping v v, e e, ( f f in E 3 ) of A onto B which preserves the incidences. But first about equivalence: A B C D E G F A B C D E G F

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Hexahedron (Cube) Octahedron Set A of vertices, edges and polygons in E 3 is combinatorially dual to such a set B if there is a 1-1 mapping v f, e e, f v of A onto B which preserves the incidences. Duality in E 3 A B c * c A B c C Voronoi Diagram (duality cont.)

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One way to perform duality is the mapping called polarity. The polarity with respect to the unit sphere centered at (0,0,0) is: pointplane This polarity maps the vertices ( 1, 1, 1) and faces x,y,z= 1 of the the cube to the faces x y z=1 and vertices ( 1,0,0), (0, 1,0), (0,0, 1) of the octahedron (previous picture). Voronoi Diagram (duality cont.)

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A B O C The polarity wrt. unit sphere centered at (0,0,0) is: Voronoi Diagram (polarity cont.) OC OA = 1 (= r 2 ) or OTA = 90 o T But for our purpose (Voronoi Diagram problem) the paraboloid is better!

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Polarity f P wrt. the paraboloid P : x 2 + y 2 = 2 z is: f P Voronoi Diagram (polarity cont.) Z (0,0,0) A

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Polarity f P preserves incidences. It maps: points of a plane to the planes through the point f P ( ), (Exercise 11 ) points of a line x to the planes through a line (def.) f P (x), (an edge AB of a polyhedron to the edge f P (A) f P (B)) ( Exercise 12 ) points of P to the planes tangent to P (Exercise 13 )... Polarity wrt. Paraboloid Exercises

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Voronoi Diagram (cont.) 2.3 First solution of the Voronoi Diagram problem Step 1. The input is a finite set S (e.g. {A,B,C,D} ) of points in a plane for which we need to construct the Voronoi Diagram = output. We are given some coordinate system in or we chose one, and add a z axis through its origin. Step 2. We project vertically (mapping denoted ) the input set S into a set S = (S ) of the paraboloid P: x 2 + y 2 = 2 z. The Voronoi Diagram is mapped onto some regions of P. Mapping has the properties:

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Z A = Voronoi Diagram (First solution cont.) M M M ~ d = d(M,A) d (M,M) = d 2 / 2 (Exercise) ~

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The points of a circle k: (x-a) 2 + (x-b) 2 = r 2 in are mapped (projected by ) to the points of some plane . (Exercise 14a) The image of an interior point of k is below = (k). (Exercise 14b) If the distance d(M,A) equals d, than the distance between M= (M) and M=MM f P (A), A= (A), equals d 2 / 2. (Exercise 15) This property has two important consequences: ~ Voronoi Diagram (Exercises )

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A B C D (1)The points M C(A) (the points closer to A than to the other points of S) have therefore the property: the plane = f P (A) is in z direction closer to M = (M), than any other such a plane f P (B), B S. (2)The points of equally distant to A and B are the vertical projection of the points f P (A) f P (B). These two properties approve the following Algorithm4. M C(A) Voronoi Diagram (First solution cont.)

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I The first step of the algorithm is the projection of the input set S onto the paraboloid. If we by A denote the projection of A, then its coordinates are: A = II In the second step we find the planes which are polar images () of these points with respect to P. These planes form a geometrical structure called arrangement of (hyper) planes*. * The polar images f P (A) of points A S, form many space regions (the cells of the arrangement). The faces of the topmost cell T are projected onto the faces of the Voronoi Diagram of S. f P Voronoi Diagram (First solution cont.) Algorithm4. = Exercise 16

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= f P ( A ) : D A C B D A C B T

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III.1 Calculate the vertices of the arrangement (find the intersections f P (X) f P (Y) f P (Z), X,Y,Z S). III.2 Find out which of these vertices are “visible from above”, i.e find out which of these vertices are not below some polar image of a point from S. These vertices are (“finite”) vertices of T, the others are thrown away. Now the “infinite” vertices: III.3 Add a new plane which closes T from above. This can be any horizontal plane : z =t which is above the vertices. Plane intersects the planes of the arrangement in “infinite” vertices of T. Faces of the polyhedron T’, made out of “finite” and infinite” vertices of T, are projected onto the Voronoi Diagram of S. III In the third step we determine the topmost cell T of the arrangement: Voronoi Diagram (First solution cont.)

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A B D C f P (A B D) f P (A B C) 2.4 The second solution of the Voronoi Diagram problem We may solve first the dual problem. The faces of the convex hull Conv (S) = A B C D..., which are “visible from below”, are mapped by polarity onto the faces of the topmost cell T of the arrangement*, and further onto the faces of the Voronoi Diagram of S. f P (C B D) f P (A C D) * hides A iff f p (A) hides f p ( ) Voronoi Diagram (Second solution)

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Voronoi Diagram (Second solution cont.) Algorithm5. (Implementation = Exercise 17 ) I As in Algorithm4. II Convex hull of S and its faces “visible from below” are computed. Note. All points of S are the vertices of Conv (S) ! III We find the polar images of the facets (2-dim. faces) of Conv (S) and project them vertically onto the plane . “visible from bellow” facets correspond to the vertices of the solution. “visible from bellow” edges correspond to the edges of the solution. an edge of Conv (S), which is on the boundary of visibility, corresponds to a half line*: starting at the image of the visible facet (incident to the edge), and not containing the image of the invisible facet. *The line of this half line, by duality, contains images of both neighboring facets!

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