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**3. Delaunay triangulation**

3.1 Euler’s formula If a planar connected graph has e edges, f facets and v vertices, than Euler’s formula states: f - e + v = 1. Similarly, for the numbers of edges, facets and vertices of a simply connected * (“sphere like”) 3-dim. polyhedron holds: f - e + v = (eq. 1) *Every simple closed polygon p, consisting of its edges, disconnects its surface: there are 2 facets of P such that every chain of adjacent facets, connecting these two, crosses polygon p. p f2 Sphere like p f2 f1 f1

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**Delaunay triangulation (Euler’s formula cont.)**

p “Donut like” By a theorem of Steiner, every graph having the properties - every facet is a polygon, - every edge is incident to exactly two facets, every vertex figure is a polygon, eq. 1 is satisfied, can be realized as a convex polyhedron. This is therefore also true for its dual graph. This duality can be realized as polarity.

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**Delaunay triangulation (Euler’s formula cont.)**

Euler’s formula gives rise to many relations between the numbers e, f and v. By the theorem of Steiner and the duality principle, every such relation holds also for the numbers e, v and f . Every facet of a polyhedron P has at least 3 edges. Summing along facets, total number of edges is 3f . In this sum every edge is counted twice (for two facets incident to it). Hence: 2 e 3 f (1), and dually 2 e 3 v (2). The inequality (1) and Euler’s formula imply: f 2 v (3), and dually v 2 f (4). The inequality (3) and Euler’s formula imply: e 3 v (5), and dually e 3 f (6)...

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**Delaunay triangulation (Polygon)**

3.2 Polygon: interior & exterior points, diagonals. Every point X, not on the boundary of a simple polygon P, has the property*: Half lines through X not through a vertex of P intersect the edges of P in an odd number of points, when X is said to be an (interior) point of P, or in an even number of points, when X is said to be an exterior point of P. • Every polygon P has inner diagonal (having only inner points). • An inner diagonal A i A k of P = A1A2…An divides the interior of P into the interiors of the polygons P1 = A i A k A k+1 … and P2 = A i A k A k-1 … *This property, as well as many polygonal properties are not easy to prove!

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**Delaunay triangulation (Triangulation)**

A k+1 A k-1 3.3 Triangulation I A k E A i A consequence of the last two properties is that every simple planar polygon admits a triangulation by its inner diagonals. If the number of triangles is t, summing the edges by triangles we get 3t=2d + n. By Euler’s formula is n - (n + d) + t = 1, hence t = n and d = n - 3.

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**Delaunay triangulation (Triangulation)**

Hence, at least two triangles of the triangulation have two sides of P as edges (each divide P into triangle and (n - 1) – gon), otherwise the number of sides would have been n-1. This property is frequently used in inductive proofs: Problem. Prove that the vertices of a triangulation of a polygon P can be colored in 3 colors so that the vertices of every triangle have different colors. More generally: A triangulation of a planar graph is its maximal, planar edge-refinement. (Faces are triangles (1), no edge is free (2), the boundary is a polygon (3), all vertices are used (4)) triangulation

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**Delaunay triangulation (Triangulation)**

If such a triangulation has n vertices on the boundary and k vertices in the interior, f triangles and e edges, a simple counting + Euler’s formula shows: f = 2 n k and e = 3 n k. In some applications, e.g. in computer graphics, is important that the triangles of a triangulation are not slim. Otherwise a human eye will see them as dashes. To get the best result we find the triangulation for which the angles of triangles are the greatest possible. This requirement leads to the following comparison:

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**Delaunay triangulation (Definition)**

Let 1 , 2 , …, 3f be the ordered sequence of angles of all the triangles of the triangulation 1 , i.e. let 1 2 … 3f Let 1 , 2 , …, 3f is another such sequence corresponding to some triangulation 2. We define ordering: 1 2 iff the corresponding sequences of numbers i and i are equal, or k < k at the first index k where they disagree. A Delaunay triangulation of a planar set of points S is its triangulation which maximizes the previous order relation.

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**Delaunay triangulation (Delaunay Graph)**

3.4 Delaunay Graph and Voronoi Diagram Voronoi Diagram of a finite planar point set S defines the following unique dual graph, named Delaunay Graph: Two points of S are joined by an edge if they define neighboring facets of the Voronoi Diagram of S. A B o D C Voronoi Diagram Delaunay Graph

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**Delaunay triangulation (Delaunay Graph)**

Let ABCD be a cell of the Delaunay Graph. Border between neighboring Voronoi Cells CA ,CB, A,B S (of the Voronoi Diagram) consists of points equally distant to A and B, and to which the other points of S are farther. It is therefore on the bisector of [AB]. Hence, the vertex o of the Voronoi Diagram common to neighb. cells of A, B, C and D is equally distant to A, B, C, D and other points of S are farther: circumscribed circle of ABCD (a facet of the Delaunay Graph) contains no other point of S. Opposite ? (Exercise) This leads to: A B o D C A subset of S defines a facet of the Delaunay Graph iff the points of S are on a circle which contains (on and in it) no other point of S.

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**Delaunay triangulation (Delaunay Graph)**

3.5 Delaunay Triangulation and Delaunay Graph We first give an algorithm which leads to a locally maximal triangulation (a). Then we prove that this triangulation is a triangulation of the Delaunay Graph of S (b). Let ABC, BCD be a pair of adjacent triangles of a triang. T. If D is interior to the circumscribed circle of ABC, then the substitution (flip) ABC,ACD ABD,BDC leads to the triangulation T’ with T T’ ()* ACB = AXB < ADB, Y C CAB = CXB < CDB, X D CAD < CAX = CBD, ACD < ACX = ABD, B *Among angles along diagonal is the smallest angle of a triangulation. Since for each such angle of T, their is a smaller one in T’, the inequality () holds. A

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**Delaunay triangulation (Delaunay Graph)**

(b) Iterating (a) we end up with a triangulation T* where all circumscribed circles are with “empty” interior. Such a triangulation T* is (see 3.4) a triangulation of the Delaunay Graph of S. Hence: Delaunay Triangulation is a refinement of the Delaunay Graph If all facets of the Delaunay Graph are triangles, i.e. if the initial set S is in general position (as we are going to assume further), the algorithm is finished. If a facet of a Delaunay Graph is a quadrangle, we easily optimize its triangulation : d 1 How to triangulate a 5-gon? - Is their an algorithm for a general problem? d1 d < d1 < 1

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**Delaunay triangulation (Algorithm6)**

Back to the algorithm (case: all facets of the Delaunay Graph are triangles. By 2.3 Exercise 14a and 3.4, the facets of the Delaunay Graph (circles …) are projected by onto some facets of the Conv (S). By 2.3 Exercise 14b and 3.4, these facets are the lower (wrt z - axis) facets of Conv (S). We are ready for the Algorithm6. Algorithm6. (Implementation = Exercise 18 ). Step1=Step1 of the algorithm4. Step 2. We calculate the lower “cap” of Conv (S). Step 3. The Delaunay Graph of S is the vertical projection of the lower cap of Conv (S) onto .

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**Delaunay triangulation (Algorithm7)**

The idea described in 3.5a (local improvements alg.) leads to: Algorithm 7. (Implementation = Exercise 19 ). Step 1. The initial set S is surrounded with a sufficiently large triangle which is the first triangle of the triangulation. Step i+1.1 New point X of S is located ( § 4) in a triangle Ti of the triangulation obtained in Step i. It separates Ti in 3 triangles. The pairs of triangles to start local improvement (diagonal flip in the corresponding 4-angle) are these 3 triangles with their neighbors! Is that all? ? Ti X Tk ? X Tk ? OK

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