Would there ever be a case where the cost of creating the KD-Tree will actually be greater than the cost of ray intersection? Will the KD-Tree ever get large enough where the cost of time or memory of it not be worth using it?
elizabethyli
I think the example that Prof. Ng walks through in the next few slides actually shows this. Someone mentioned that it seems like there were more intersection checks that happened when we used the tree compared to the naive techniques, meaning the space is over-partitioned or the partitions were badly chosen. The technique is much more effective when there's a lot of objects.
Would there ever be a case where the cost of creating the KD-Tree will actually be greater than the cost of ray intersection? Will the KD-Tree ever get large enough where the cost of time or memory of it not be worth using it?
I think the example that Prof. Ng walks through in the next few slides actually shows this. Someone mentioned that it seems like there were more intersection checks that happened when we used the tree compared to the naive techniques, meaning the space is over-partitioned or the partitions were badly chosen. The technique is much more effective when there's a lot of objects.
If choose the midpoint, is KD-Tree also OCT-Tree?