# COSC 6114 Prof. Andy Mirzaian Quad Trees: Non-Uniform Mesh Generation.

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COSC 6114 Prof. Andy Mirzaian Quad Trees: Non-Uniform Mesh Generation

References: Applications: [M. de Berge et al] chapter 14
Marshal Bern [1997], “Triangulations,” in Handbook of Discrete and Computational Geometry, J.E. Goodman & J. O’Rourke (editors), chapter 22, Applications: VLSI circuits in electrical devices heat emission – a design issue experiment with prototype automated simulation by finite-element methods MESH (uniform / non-uniform)

Applications & Methodology: VLSI circuits in electrical devices heat emission – a design issue experiment with prototype; reject faulty design (expensive, time consuming) automated simulation by finite-element methods subdivide regions into sufficiently fine triangles or quadrangles: MESH (uniform/non-uniform) what is the influence of one mesh element on neighboring elements? mesh should be more refined near the “objects” and gracefully get coarser further away avoid mesh elements that have extreme aspect ratios (bounded angles) mesh elements should respect object borders.

Uniform vs Non-Uniform Mesh
Sufficiently refined. Too many mesh elements (time & space) U = 2j integer grid U U

Uniform vs Non-Uniform Mesh
Conforming Respect Input Well Shaped: bounded aspect ratio (e.g., angels  [45 : 90]) Steiner Points

Uniform vs Non-Uniform Mesh
Unifrom Mesh too many triangles

Uniform vs Non-Uniform Mesh
based on: Constrained Delaunay Triangulation small angels

Uniform vs Non-Uniform Mesh

NW 2 NE 1 SW 3 SE 4 1 NE 2 NW 3 SW 4 SE

22 21 1 4 1 2 3 242 241 23 243 244 24 21 22 23 31 32 33 34 31 32 4 241 242 243 244 33 34

square/node: s Point set P  [xs : x’s ]  [ys : y’s ] xmid = (xs + x’s )/2 , ymid = (ys + y’s )/2 PNE = { p P | px > xmid , py > ymid } PNW = { p P | px  xmid , py > ymid } PSW = { p P | px  xmid , py  ymid } PSE = { p P | px > xmid , py  ymid } Keep splitting a square if it contains more than one data point. edge side corner

LEMMA: P = a set of points in the plane,
LEMMA: P = a set of points in the plane, S = side length of root square for Quad Tree of P, C = smallest distance between pair of points in P, D = depth of Quad Tree. Then, D  log S/C + 3/2 . Proof: Any internal node contains at least 2 points of P. Node at depth i has side-length S/2i , diagonal length (S2)/2i  C So, i  log (S2)/C = log S/C + ½ . Deepest leaf has depth one more.

Balanced Quad Trees Scene gives un-balanced Quad Tree neighbor squares with un-balanced sides resulting mesh has tiny angles To achieve conformity and satisfy angle-bounds (aspect ratio), We need to balance the QuadTree. Balanced QuadTree: side-length ratio of any 2 neighboring squares = O(1).

Balanced Quad Trees Two squares are neighbors if they have overlapping sides but disjoint interiors. Quad Tree is balanced if any two leaf neighboring squares differ in side-length by at most 2, i.e, have depth difference  1.

Balanced Quad Trees Two squares are neighbors if they have overlapping sides but disjoint interiors. Quad Tree is balanced if any two leaf neighboring squares differ in side-length by at most 2, i.e, have depth difference  1. BALANCING

Balanced Quad Trees s = a leaf of Quad Tree T NN(s) WN(s)
NN(s) = North Neighbor of s, is a node s’ in T s.t. (i) north edge of s is shared by s’ (ii) depth(s’)  depth(s) (i.e., square s’ not smaller than s) (iii) s’ is the smallest such square in T. SN(s), EN(s), WN(s) defined similarly. NN(s) WN(s) EN(s) s SN(s) SN(s) EN(s) s NN(s) WN(s)

Balanced Quad Trees These take O(D) time, D = height(T ).
LEMMA: Given a leaf s in T , NN(s), NN(s), NN(s), NN(s) can be obtained in O( depth(s)). Proof: Let s’ = NN(s). From s, go up in T to lowest common ancestor of s & s’, then come down to s’. LEMMA: Leaf s should split   leaf s’ in T , s.t. s is one of NN(s’), NN(s’), NN(s’), or NN(s’), & depth(s’)  2 + depth(s). These take O(D) time, D = height(T ).

Balancing a Quad Tree while L   do if s has to split then do
ALGORITHM BalancedQuadTree(T ) Input: Quad Tree T Output: a balanced version of T L  list of all leaves of T while L   do remove a leaf s from L if s has to split then do add 4 children sNE , sNW , sSE , sSW to s in T & update their object contents insert sNE , sNW , sSE , sSW into L check if sNE , sNW , sSE , sSW have neighbors that should split & add them to L end-if end-while return T end

THEOREM: Let T be a quad-tree with m nodes and height D
THEOREM: Let T be a quad-tree with m nodes and height D. Then, the above algorithm constructs a balanced version of T that has O(m) nodes in O(Dm) time.

THEOREM: Let T be a quad-tree with m nodes and height D
THEOREM: Let T be a quad-tree with m nodes and height D. Then, the above algorithm constructs a balanced version of T that has O(m) nodes in O(Dm) time. Proof: Step 1: takes O(m) time. Steps 2-9: each iteration adds 4-1 = 3 nodes to T & takes O(D) time to check O(1) neighboring nodes in L . Therefore, total time for the while-loop is O(D  # new nodes added). What is the number of added new nodes?

THEOREM: Let T be a quad-tree with m nodes and height D
THEOREM: Let T be a quad-tree with m nodes and height D. Then, the above algorithm constructs a balanced version of T that has O(m) nodes in O(Dm) time. Proof: Step 1: takes O(m) time. Steps 2-9: each iteration adds 4-1 = 3 nodes to T & takes O(D) time to check O(1) neighboring nodes in L . Therefore, total time for the while-loop is O(D  # new nodes added). What is the number of added new nodes? Even though some leaf in s1 causes leaf s2 to split, that splitting cannot propagate to the neighbor s3 of the same size. [Proof by induction on node depth difference.] NW NE SW SE s1 s2 s3

THEOREM: Let T be a quad-tree with m nodes and height D
THEOREM: Let T be a quad-tree with m nodes and height D. Then, the above algorithm constructs a balanced version of T that has O(m) nodes in O(Dm) time. Proof: Step 1: takes O(m) time. Steps 2-9: each iteration adds 4-1 = 3 nodes to T & takes O(D) time to check O(1) neighboring nodes in L . Therefore, total time for the while-loop is O(D  # new nodes added). What is the number of added new nodes? Even though some leaf in s1 causes leaf s2 to split, that splitting cannot propagate to the neighbor s3 of the same size. [Proof by induction on node depth difference.] NW NE SW SE s1 s2 s3 charge 1 to neighbor of equal size that causes splitting each square is charged  8 times. range of influence:

Root square contains polygonal objects, all vertices at integer grid points (U=2j), all edges are at angels 0, 45, 90, 135. Stop splitting when the square no longer intersects with any object edge, or when it has unit size. Balance the resulting Quad Tree for each leaf square s do: if there is a diagonal object edge, fine; otherwise, add one such diagonal to make s conform to a neighboring smaller leaf, add O(1) horizontal or vertical or diagonal lines also. all angels 45 or 90.

Then, there is a non-uniform triangular mesh for S that is:
THEOREM: Let S = a set of disjoint polygonal components inside the square [0 : U][0 : U] with properties stated earlier, & p(S) = total perimeter length of components in S. Then, there is a non-uniform triangular mesh for S that is: conforming, respects the input, triangle angles are 45 or 90, # triangles = O(p(S) log U), Construction time = O(p(S) log2 U). Proof: # triangles incident to an input segment of length L is  2(L+2). Thus, # cells in the quad-tree at the same depth is O(p(S)). Depth of quad-tree is at most O(log U). Thus, total # nodes in quad-tree is O(p(S) log U). The rest follows from previous discussion: m = O(p(S) log U) D = O(log U) O(Dm) = O(p(S) log2 U).

Notes & Comments [Bern 1997]: any polygonal domain with n vertices & no obtuse angles has a mesh consisting of O(n) non-obtuse triangles (i.e., all angels  90). Other applications of quad-trees (oct-trees in 3): computer graphics image analysis range queries hidden surface removal ray tracing medial axis transforms overlay of raster maps nearest neighbor query processing   

Exercises

Suppose a triangular mesh is needed inside a rectangle whose sides have length 1 and length k > 1. Steiner points may not be used on the sides, but they may be used inside the rectangles. Also assume that all triangles msut have angles between 30 and 90. Is it always possible to create a triangular mesh with these properties? Suppose it is possible to create a mesh for a particular input, what is the minimum number of Steiner points needed? The algorithm described produces non-obtuse triangulated mesh (provided all angles in the input have no obtuse angles). Prove that if a triangulation of a set P of points in the plane contains only non-obtuse triangles, then it must be the Delaunay triangulation of P. Describe an algorithm to construct an oct-tree of a given set P of n points in 3D. It is possible to reduce the size of a quad-tree of height D for a set of points (with real coordinates) inside a square from O((D+1)n) to D(n). The idea is to discard any node v that has only one child under which points are stored. The node is discarded by replacing the pointer from the parent of v to v with the pointer from parent to the only interesting child of v. Prove that the resulting tree has linear size. Can you also improve upon the O((D+1)n) construction time? We called a quad-tree balanced if every two adjacent squares of the quad-tree subdivision differ by no more that a factor of two in size. To save a constant factor in the number of extra nodes needed to balance a quad-tree, we could weaken the balance condition by allowing adjacent squares to differ by a factor of four in size. Can you still complete such a weakly balanced quad-tree subdivision to a mesh such that all angles are between 45 and 90 by using only O(1) triangles per square?