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Solutions to homework #4 and #5 Wu Xiaomao May 5, 2004.

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Presentation on theme: "Solutions to homework #4 and #5 Wu Xiaomao May 5, 2004."— Presentation transcript:

1 Solutions to homework #4 and #5 Wu Xiaomao May 5, 2004

2  12.1. Define the results of performing and for two polyhedral objects in the same way as the result of performing was defined in section 12.2. Explain how the resulting obje- cts is constrained to be a regular set, and sp- ecify how the normal is determined for each of the object’s faces.

3  Solution:  The union operation is defined as: A B=closure(interior(A B)) A B A B A B A B A B A B NaNa NbNb NbNb NaNa NaNa NbNb

4 The union operation is defined as: A B=closure(interior(A B)) A B NaNa NbNb A B N A B NbNb NaNa A B A B NbNb NaNa A B

5 9.13 Explain why an implementation of Boo- lean set operations on quadtrees or on octr- ees does not need to address the distinction between the ordinary and regularized opera- tions described in Section 12.2

6 Solution: The reason why we need to distinguish between ordinary and regularized Boolen set operations is that we want to resulting objects to be a regularized objects( no dangling edges or dangling surfaces). In quadtrees or octrees representation, objects are expressed in a discrete way. Here no boundary or surface concept is concerned. The Boolean set operation is done by compare the nodes one by one.

7 12.8 Describe how to perform point classifi- cation in primitive instancing, b-rep, spatial occupancy enumeration, and CSG

8 Solution: 1.Primitive instancing: As primitive instance is created by codes, the programmer should design a subroutine for point classification and embed it in the primitive instance. The user can load these subroutines when he/she what to do a point classification.

9 2. b-rep Odd intersection points: In Even intersection points: Out A B C D E F 1 2 3 4

10 3. Spatial occupancy enumeration Search the quadtree/octree, if point A belongs to one of the empty node, then point A is outside. If point A belongs to one of the full node, continue searching the neighbors of point A (minimum unit). A is on the boundary if one of A’s neighbor is a empty node. A is inside if all its neighbors are full nodes or within a full node.

11 4. CSG Determine whether the point is inside/outside/ on the boundary for each primitive. Let that point join the Boolean set operation to determine its relation to the object.

12 Supplemental problem: V=16 E=24 F=11 V-E+F=3 V=16 E=28 F=14 V-E+F=2

13 Supplemental problem: V=16 E=24 F=11 H=1 C=1 G=0 V-E+F-H=2 2(C-G)=2 V-E+F-H=2(C-G)

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