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**Delaunay Triangulation and Tetrahedrilization**

Marc van Kreveld Slides DDM

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**Triangulations and their quality**

When going from points on a surface to triangles representing that surface, there are many ways to form the triangles Some ways are better than others

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**Triangulations and their quality**

This is not just true visually, but also when the triangulation is the interpolator for a terrain with elevation measurements

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**Triangulations and their quality**

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**Triangulations and their quality**

Triangulations are good if they do not have long edges do not have triangles with both short and long edges do not have triangles with very small angles do not have triangles with obtuse angles do not have vertices with high degree …

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Voronoi diagrams For a set of points (sites), the Voronoi diagram is the subdivision of the plane (space) with faces where exactly one of the sites is closest, among all sites

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Voronoi diagrams Voronoi vertices “often” have degree 3, but it can be much larger Voronoi edges can be bounded or unbounded Voronoi edge Voronoi vertex

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Voronoi diagrams With n point sites, there are < 2n Voronoi vertices and < 3n Voronoi edges Voronoi edge Voronoi vertex

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Voronoi diagrams The average number of edges bounding a Voronoi cell is six (or slightly less), but a single cell could be bounded by up to n – 1 edges Voronoi edge Voronoi vertex

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Voronoi diagrams Voronoi diagrams have the empty circle property for every Voronoi vertex and Voronoi edge

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Voronoi diagrams Efficient algorithms exist to construct 2D Voronoi diagrams of n point sites, namely O(n log n) time algorithms (as fast as sorting! [mergesort, quicksort])

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Voronoi diagrams 3D Voronoi diagrams with n point sites can have up to (n2) Voronoi vertices (0D), Voronoi edges (1D), and Voronoi facets (2D) but of course just n cells (3D) The descriptive complexity can be anything between linear and quadratic in n

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Voronoi diagrams 3D Voronoi diagrams with n point sites often have linearly many Voronoi vertices (0D), Voronoi edges (1D), and Voronoi facets (2D)

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**Delaunay triangulation**

For a set of points in the plane, suppose we compute its Voronoi diagram and use it to make an embedded planar graph G as follows: all points are vertices of the graph G for all points whose Voronoi cells share a Voronoi edge, we have an edge in G faces of G are implied by the vertices and edges of G

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**Delaunay triangulation**

all points are vertices of the graph G edges in G correspond to edge-adjacent Voronoi cells faces of G are implied by the vertices and edges of G

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**Delaunay triangulation**

all points are vertices of the graph G edges in G correspond to edge-adjacent Voronoi cells faces of G are implied by the vertices and edges of G

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**Delaunay triangulation**

This graph is called the Delaunay graph Notice that an edge of the Delaunay graph does not necessarily intersect the corresponding Voronoi edge

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**Delaunay triangulation**

The Delaunay graph is a triangulation of the original points if and only if all Voronoi vertices have degree 3 there is no circle through more than 3 sites and with no sites inside

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**Delaunay triangulation**

Any (planar) completion of the Delaunay graph to a triangulation is a Delaunay triangulation of the original points

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**Delaunay triangulation**

The empty-circle property for Voronoi diagrams transfers to Delaunay triangulations for each triangle, its circumcircle contains no other points from the set inside for each edge, a circle exists through its endpoints that has no other points of the set inside

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**Delaunay triangulation**

A Delaunay triangulation of n points has at least n – 2 and at most 2n – 5 triangles (Delaunay triangles) A Delaunay triangulation of n points has at least 2n – 3 and at most 3n – 6 edges (Delaunay edges) The empty-circle property is “if and only if”: if three points have a circumcircle with no points inside or other points on the boundary, then they make a Delaunay triangle (a similar statement holds for edges)

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**Delaunay triangulation**

When a Delaunay graph is not a triangulation, the non-triangular faces have all vertices on a circle

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**Delaunay triangulation**

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**Delaunay triangulation**

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**Delaunay triangulation**

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**Delaunay triangulation**

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**Delaunay triangulation**

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**Delaunay triangulation**

When we know four Delaunay edges that form an empty convex quadrilateral, we will have one of the two diagonals as a Delaunay edge The circumcircles of the triangles of one choice will contain the fourth points, but not for the other choice

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**Delaunay triangulation**

The Delaunay triangulation maximizes the smallest occurring angle, over all triangulations of the point set In other words, any other triangulation will have a smaller (or the same) smallest angle

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**Delaunay triangulation**

The Delaunay triangulation often connects points close together, e.g., every point is connected to its nearest neighbor But it does not minimize total edge length

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**Delaunay triangulation**

In a triangulation, when two triangles form a convex quadrilateral, their shared edge can be replaced by one other edge This is called an edge flip flip

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**Computing the Delaunay triangulation**

Approach: insert points one-by-one, and restore the Delaunay triangulation after every insertion Locate the triangle t con-taining the next point p Connect p to t’s vertices Restore the Delaunay triangulation by flipping non-Delaunay edges p t

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**Computing the Delaunay triangulation**

Locate: Recall that we store triangulations in some convenient structure (triangle-neighbor structure, half-edge structure) From any access point (triangle, half-edge), walk along a straight line to the location of the p, finding the triangle t containing p The details of the walk depend on the mesh structure used p

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**Computing the Delaunay triangulation**

Connect: Make edges from p to t’s vertices; this replaces t by three new triangles Adapt the mesh representation structure accordingly Claim: These three new edges are Delaunay p

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**Computing the Delaunay triangulation**

Proof of claim: Triangle t was Delaunay before the insertion of p, so there was an empty circle C through its vertices C v The growing circle starts as a single point equal to v, and then the growing circle has its center move in a straight line to the center of C, while keeping passing through v. This keeps the growing circle tangent to C at v. No other point than p can be hit in the growing process, because we know that p is the only point inside C. p t

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**Computing the Delaunay triangulation**

Proof of claim: Triangle t was Delaunay before the insertion of p, so there was an empty circle C through its vertices Consider the edge between p and any vertex v of t, there is always a circle through p and v completely inside C (grow a small circle from v tangent to C at v, until it hits p) C v The growing circle starts as a single point equal to v, and then the growing circle has its center move in a straight line to the center of C, while keeping passing through v. This keeps the growing circle tangent to C at v. No other point than p can be hit in the growing process, because we know that p is the only point inside C. p t

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**Computing the Delaunay triangulation**

Proof of claim: Triangle t was Delaunay before the insertion of p, so there was an empty circle C through its vertices Consider the edge between p and any vertex v of t, there is always a circle through p and v completely inside C (grow a small circle from v tangent to C at v, until it hits p) Since v and p have an empty circle, they define a Delaunay edge C v The growing circle starts as a single point equal to v, and then the growing circle has its center move in a straight line to the center of C, while keeping passing through v. This keeps the growing circle tangent to C at v. No other point than p can be hit in the growing process, because we know that p is the only point inside C. p t

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**Computing the Delaunay triangulation**

Restore: The three new edges are always Delaunay, but the three new triangles need not be … Delaunay p p maybe not Delaunay

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**Computing the Delaunay triangulation**

In the picture: triangle qrs had an empty circle C(q,r,s) before the insertion of p, but maybe p lies inside now test p C(q,r,s), and if so, edge-flip qr to ps Delaunay q q C(q,r,s) p p s s r r maybe not Delaunay

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**Computing the Delaunay triangulation**

If flipped, the edges pq, pr, and also ps must be Delaunay, but maybe pqs and prs are not Delaunay triangles … recurse on them, using the triangles opposite qs and rs Delaunay Delaunay q q p p s s r r maybe not Delaunay maybe not Delaunay

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**Computing the Delaunay triangulation**

Code for recursive flipping (restore algorithm): TestFlipEdge (p, qr) Let s p be the third point of the triangle incident to qr if p is inside C(q,r,s) { flip: delete qr and insert ps in the triangulation TestFlipEdge(p, qs) TestFlipEdge(p, sr) }

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**Computing the Delaunay triangulation**

Comments: The edge flip must be done in our triangle-mesh representation structure (triangle-neighbor, half-edge, …) With suitable triangle-mesh structure, a flip takes O(1) time (with an indexed mesh, a flip takes O(n) time) Every edge flip connects p to an extra vertex on the average, about 3 flips are needed The main geometric test is the so-called in-circle test: does a point p lie inside the circle defined by three points q,r,s?

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The in-circle test The in-circle test: does a point p lie inside the circle defined by three points q,r,s? (1) The bad way: Compute the bisectors of q,r and r,s Intersect them to get the center c of C(q,r,s) Compute dist(c,p) and dist(c,q) (the radius of C(q,r,s)); if the former is smaller, then p lies inside C(q,r,s)

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The in-circle test The in-circle test: does a point p lie inside the circle defined by three points q,r,s? (2) The good way: Compute the plane through (qx, qy, qx2 + qy2), (rx, ry, rx2 + ry2), and (sx, sy, sx2 + sy2) Test whether the point (px, py, px2 + py2) lies above or below this plane: below p is inside; above p is outside Geometrically, we are lifting the points q,r,s from the plane onto the unit paraboloid in 3D (z=x^2+y^2). The plane through these three 3D points intersects the unit paraboloid exactly in a shape that projects back to a circle on the xy-plane.

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The in-circle test The in-circle test: does a point p lie inside the circle defined by three points q,r,s? (2) The good way: Compute the plane through (qx, qy, qx2 + qy2), (rx, ry, rx2 + ry2), and (sx, sy, sx2 + sy2) Test whether the point (px, py, px2 + py2) lies above or below this plane This is equivalent to computing the sign of the 4x4 determinant: qx qy rx ry qx2 + qy2 1 rx2 + ry2 1 sx sy px py sx2 + sy2 1 px2 + py2 1 negative inside

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The whole algorithm Initialization: we want to make sure that the next point p is always in some triangle of the current triangulation start with a suitable bounding box (triangulated)

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The whole algorithm The four extra vertices need special treatment when they are involved in an in-circle test (because they do not count for the empty-circle property)

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**The whole algorithm Initialize a triangulation T with a bounding box**

For each point pi Locate the triangle t of T containing pi by traversing the triangle-mesh structure from some access point Connect pi to the vertices of t (Restore) For each edge e of t, TestFlipEdge(p, e)

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**Efficiency of the algorithm**

Locate: if the points are distributed “reasonably”, a line typically intersects O(n) triangles (certainly true if the points lie in a regular square grid pattern) Worst-case O(n) time Recall that a triangulation with n vertices has at most 2n triangles and at most 3n edges some fixed starting point

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**Efficiency of the algorithm**

Connect: obviously O(1) time in the triangle-neighbor structure and half-edge structure Restore: typically 3 flips are needed: the average degree of a vertex in a triangulation is 6, connect leads to a starting degree of 3 for pi, and every flip increases the degree of pi by 1 typically O(1) time, but worst-case O(n) time

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**Efficiency of the algorithm**

Initialization O(n) time Insertion of the i-th point takes worst-case O(i) time but typically O(i) time Worst-case O(n2) time algorithm Typically O(nn) time algorithm Note: worst-case O(n log n) time algorithms exist; these are more complicated

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**Delaunay tetrahedrilization**

Global approach the same The in-circle test becomes an in-sphere test (solved with a 5x5 determinant) The edge-flip becomes a bi-stellar flip: 2 3 tetrahedra

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**Delaunay tetrahedrilization**

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**Terrains and 2D/3D Delaunay**

Note the difference: A polyhedral terrain (triangular mesh representing elevation) obtained by computing the Delaunay triangulation of the (x,y) points and using the triangles in 3D A 3D Delaunay tetrahedrilization of the (x,y,z) points A triangle in the first case does not necessarily occur as a triangular facet in the second case (the 2D circle can be small when the 3D ball is huge)

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**Constrained Delaunay triangulation**

Variation on the Delaunay triangulation where certain edges must be included (even if they are not Delaunay) Endpoints of such constraining edges are also vertices of the constrained Delaunay triangulation

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**Constrained Delaunay triangulation**

Useful when known linear features must be included in the triangulation some polygon boundary lake boundaries in a terrain with elevation values known

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**Constrained Delaunay triangulation**

Input: set of points and set of edges (constraints)

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**Constrained Delaunay triangulation**

Input: set of points and set of edges (constraints) constrained Delaunay triangulation

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**Constrained Delaunay triangulation**

Input: set of points and set of edges (constraints) normal Delaunay triangulation

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**Constrained Delaunay triangulation**

Relaxed empty-circle property: three points define a circle in the constrained Delaunay graph if and only if their circumcircle has no other points (including constraint endpoints) inside or on the boundary, with the possible exception of points on the other side of constraints that cut the circle fully

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**Constrained Delaunay triangulation**

To compute the constrained Delaunay triangulation, first compute the normal Delaunay triangulation of the points and constraint endpoints Then insert the constraints one by one, removing the intersected edges/triangles, and re-triangulating the holes

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**Constrained Delaunay triangulation**

Build Delaunay triangulation

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**Constrained Delaunay triangulation**

Insert constraining edge by removing intersected edges and triangles

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**Constrained Delaunay triangulation**

Note: all other edges and triangles remain in the CDT, because the empty circle of the DT will also be an empty circle with the constraint added

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**Constrained Delaunay triangulation**

Triangulate the cavity by shrinking a circle through the constraint endpoints until it reaches a vertex

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**Constrained Delaunay triangulation**

Make two new edges to this vertex from the other points defining the circle, and recurse

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**Constrained Delaunay triangulation**

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**Constrained Delaunay triangulation**

A constrained Delaunay triangulation can be constructed in O(n log n) time in the worst case (for n points and constraints) The given algorithm takes O(n3) time in the worst case but one expects much better performance

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**Constrained Delaunay in 3D**

Constrained Delaunay tetrahedrilizations do not always exist We would need to subdivide constraining facets using extra vertices and edges

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**Steiner points in triangulations**

The Delaunay triangulation of the given points may not give sufficiently well-shaped triangles add extra points (Steiner points) Useful for point sets with or without constraints

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**Steiner points in triangulations**

Steiner points can lower the maximum vertex degree and improve the smallest occurring angle Steiner points placed on a constraint can break it into several shorter constraints

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**Steiner points in triangulations**

A common example is triangulating a simple polygon with Steiner points on the edges and inside

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Questions Prove that edge ps (slide 38/39) is always Delaunay, using the empty circles known from the situation before p was inserted Why is the running time of inserting all constraints incrementally into a constrained Delaunay triangulation cubic in the number of points and constraints? Why do we expect much better performance than cubic when inserting all constraints?

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