Binary space partitioning pdf

In: Jawahar, C. ACCV Gkioxari, G. Remote Sens. He, K. Accessed 28 Jan Kingma, D. Kirillov, A. Lee, S. Long, J.

In: CVPR, pp. Loshchilov, I. Mescheder, L. Papandreou, G. Ronneberger, O. In: Navab, N. Sandler, M. Sanglard, F. Sanglard, Fabien Google Scholar. Simonyan, K. Belcour, L. ACM Trans. Graphics 37 2 , — Dupuy, J. Graphics 36 4 , — Heitz, E. Graphics 35 4 , — Ho, T. IEEE Trans. Visual Comput. Graphics 21 8 , — Iwasaki, K. Graphics Forum 31 2 , — Graphics Forum 33 2 , — Lecocq, P. Graphics 23 5 , — Ng, R. Graphics 22 3 , — Nowrouzezahrai, D.

Graphics Forum 33 1 , — Sloan, P. Graphics 21 3 , — Soler, C. Graphics 34 6 , — Sun, B. Graphics 28 2 , Tsai, Y. Graphics 25 3 , — Wang, J. Graphics 37 4 , — The result is either T's value e. That is, complemented. The tree now represents the desired set. The algorithm returns a BSP tree v. The else v. Psuedo code for the incremental set partition, we have a CSG tree with faces at its leaves.

Figure CSG- evaluation algorithm. As before, a hyperplane H is chosen at each stage that embeds a face using a heuristic Section 5. Two copies of the CSG tree are generated and modified to each primitive replacing the faces of that primitive in the respective represent the set in each of the two halfspaces of H. This entails for CSG trees with the subset of the faces that lies in each halfspace.

Detection of faces which lie in these sub-hyperplanes can possibly be in bd S. If the CSG tree is reduced to, in effect, a single superset of bd S , i. It also immediately follows represented by a cell of the BSP tree. How- ous with respect to the defined set, where each region is represented by ever, parts of these faces may lie in either int S or ext S.

To find the a different cell of the resulting BSP tree. This produces a set of new faces, a subset of which form bd S I" SHp v , and this subset is retained at v as choose a f a c e f o f a primitive of r opposed to extending the tree as in sections 3 and 4.

Consider for the moment the case where v. Moreover, the orientation of the boundary faces must be that of else if rteft represents U then Hv, since they are to point to the exterior, which by construction lies in v. Therefore, faces in the opposite-face list cannot be in bd S.

These can be return v discarded. Consider union. Therefore, we know a priori that same-faces Add Tcoincident to V'S face lists inserted into v. Algorithm for converting from a C S G cells.

For intersection, a similar analysis indicates that same-faces tree to a B S P tree. While each node are disjoint. If the faces are given the same attributes, such certain faces were retained at internal nodes, these no longer as color, this redundancy will not affect renderings of the object, other correspond necessarily to the boundary of the set S represented by the than to possibly increase time and space requirements.

However, this tree S. Since B-reps are useful for rendering via polygon tiling, and redundancy can be eliminated by merging the faces, i. This requires that for each node the union of the opposite-faces.

There are two alternatives. One is to dis- card the old faces entirely and generate the boundary faces directly 4. Conceptually, we the faces of the operands. The result is a BSP tree, in d - 1 - s p a c e , where the d - l value of The boundary of the result of any set operation is known to be a subset "in" corresponds to the d-value of "on", and similarly "out" corresponds of the boundaries of the operands.

Now, since bd S is known to lie to "not-on". The tree building process [Putn86], to "regularize" the set, proceeds a s before, but in the lower dimension. The recursion ter- minates when regions are discovered that are either completely covered 4.

If convex polygons are the tain nodes without changing the represented set. We identify two desired output, it is relatively straightforward to maintain a vertex-list cases in which this reduction is possible.

The first case occurs when representation of the regions of the 2D tree. All in-regions yield both subtrees of a node v are cells with identical values Node z in polygons whose vertices are projected back into H.

This v can be replaced by a cell with the same vatue. Note that no boun- means that finding the 2D boundary of a 3D set requires recursing in dary faces could lie in the sub-hyperplane of such a node, This case dimension and finding the 1D boundary of 2D set and subsequently the arises naturally from expression simplification during which a formerly 0D boundaries of 1D sets. It can be performed sion. The recursion forms 1D BSP trees for each internal node of a as part of the tree construction.

The in-cells of these trees lie on polygon edges. In this I D-space, hypcrplanes are forced to the form [ 1 - x ]. This means that all cells in the other subtree bounded are binary search trees. To find the minimum boundary of these 1D by this node's hyperplane have the same value as the cell. Since the in-regions, i. The vertices are projected back from tD to 2D which are tree Node w.

This reduction can be performed during the phase that then projected back into H defined in 3D. This then produces a generates the boundary faces. With the incremental algorithm, this merged set of edges bounding the on-regions with respect to the 3D can be detected and effected during set-op evaluation. Nodes u and z can be eliminated. We solve this for dimensions 1, 2 and 3. We assume, without loss of inserted into v.

The complementary operation is performed for the We are then interested in determining the status of R v. The resulting in-cell faces are retained and merged respect to B. In the case where B O R v is convex, this is rela- together as above, but with the following "glue" operator in place of tively simple. If the point is "behind" all of these faces, then R v. The I D boundaries of same and opposite regions are constructed We now address the problem for sets of arbitrary polyhedra.

One independently. This kind of operation has appeared elsewhere, e. This method would inter- sect a ray emanating from a point. R e p r e s e m i n g a set of a r b i t r a r y non-overlapping polygons by a set of edges is find the closest face, from whose orientation the classification can be sufficient for m a n y polygon processing a l g o r i t h m s. If the closest intersection point lies on more than one face, the process is repeated with a randomly perturbed ray.

This b can be found trivially during the partitioning of that select a hyperplane from among those that embed faces. In the following, l e t p be a point in S H p v. The test set con- orientation of the single face a point. Vertex b is either in bd R v or in with possible outcomes being "in front oP', "in back of", and "inter- int R v.

If b lies in b d R v , then there is a single edge e in Bv sected by". The heuristic is a function of the number of outcomes of incident with b. A second edge could only lie in b d R v or each type that occurred when a candidate was tested against the test ext R v. Otherwise, set. The candidate chosen is that member of the candidate set that R v.

Now, if b is in int R v , b is incident with two maximizes the heuristic. We investigated three heuristic functions: edges, el and e2. The region R v.

Otherwise R v. The weight wsptit allows "tuning" of the heuristics. The reason for applying a negative weight to intersected candidates is that splitting of tt". This can facilitate CSG tree simplification, since in one of the two subspaces, the value of the primitive will be out.

Pascal running under Unix BSD 4. The CSG tree is described in a simple language of our design, translated using lex and yacc. Tests were run on a of the single face containing b. When b is shared by more than one VAX If f l and f 2 are problem.

Early experience with various candidate set sizes shows that the faces that share this edge, then R v.

Heur2 produces trees region of b, f l and f 2 lie in each other's back halfspace; otherwise, with a larger number of nodes, but with less CPU time than is R v.

To determine this we first find a vertex of f l required by the other heuristics for the same objects. If the faces are convex, any vertex of We have implemented the algorithm for incremental set operations in f l not lying in f 2 will do.